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math4finance
x=a) 27b) 54c) 61
Accepted Solution
A:
Answer:The measure of ∠x=27° Step-by-step explanation:Given the figure we have to find the value of x As the arc AC is 68° i.e m arc(AC)=68°By theorem, the angle subtended at the centre is twice the angle subtended at the circumference of circle i.em arc(AC)=2∠ABC[tex]\angle ABC=\frac{1}{2}m arc(AC)=\frac{1}{2}\times 68=34^{\circ}[/tex]As ∠BED=61°∠AEB and ∠BED forms a linear pair therefore their sum adds up to 180°∠AEB + ∠BED=180°∠AEB =180°-61°=119°In ΔAEB, by angle sum property∠BAE+∠ABE+∠AEB=180°x+34°+119°=180°x=180°-34°-119°=27°Hence, the measure of ∠x=27°Option a is correct
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Related Puzzles
Finish the Lyrics
Making of America
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Business Communication
Layers of the Ocean
QUESTIONS LIST: line: a straight one-dimensional figure that has no thickness and extends in both directions, segment: line _ is a piece or part of a line having two endpoints. , intersecting lines: the crossing of two lines, ray: begins at one end but goes on forever in the other direction. , parallel: _ lines that never cross, right: _ angle is an angle of 90 degrees, acute: _ angle is an angle less than 90 degrees, obtuse: _ angle is an angle greater than 90 degrees, complementary: two angles with a sum of 90 degrees, supplementary: angles that add up to 180 degrees, polygon: a closed two-dimensional shape such as - rectangle or triangle, , quadrilateral: a four-sided figure, congruent: having the same shape and size, also described as the mirror image of the other, coordinate: a grid formed by a horizontal and vertical number line is called _ plan, ordered: _ pair is a pair of numbers of (x,y) that is used to describe a location on a coordinate plane
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Hello there, geometry enthusiasts! Are you ready to dive into the thrilling world of triangle vertices calculation? Buckle up for an exciting journey through the land of angles and vertices. Just a heads up though, don't take any acute angles to heart!
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1. A 23-tooth helical gear is cut with a 20° pressure-angle hob at a helix angle of 25°. The hob has a standard diametral pitch of 5. The resulting teeth have standard spur-gear dimensions in the normal plane. Find the pitch diameter, addendum, dedendum, outside diameter, normal, transverse, and axial pitch, and the transverse pressure angle.
2. A 38-tooth helical gear is cut with a 25° pressure-angle hob at a helix angle of 30°. The hob has a standard diametral pitch of 4. The resulting teeth have standard spur-gear dimensions in the normal plane. Find the pitch diameter, addendum, dedendum, outside diameter, normal, transverse, and axial pitch, and the transverse pressure angle
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RD Sharma Class 11 Chapter 5
RD Sharma Class 11 Chapter 5 Trigonometric Functions is an important chapter for students and they should not miss it. It contains a lot of formulas and can be quite tricky to learn but at Vedantu, we provide simpler methods which are easy to understand for the students to get the concept and score good marks in exams. This chapter requires a lot of focus and practice because of the number of formulas that are there.Trigonometric functions are also known as angle functions and are developed to solve problems regarding triangles. The functions used are Sin, Cos, tan and their reciprocal identities are cosecant, secant and cotangent. Students will come across a lot of different identities in this chapter which they will need to learn because it is needed in their higher class as well.
RD Sharma Solutions for Class 11 chapter 5 is provided in easy and understandable terms for better understanding of students who are writing their boards and other competitive exams soon. It is provided with a step by step method keeping in mind that each student has a different level of understanding.
Class 11th Chapter Trigonometric Function Exercises and Questions Covered in NCERT Textbook along with the Number of Questions Covered
Exercise 3.1- 7 Questions
Exercise 3.2- 10 Questions
Exercise 3.3- 25 Questions
Exercise 3.4- 9 Questions
Miscellaneous Exercise- 10 Questions
List of Basic Trigonometric Functions
(image will be uploaded soon)
Sin0 = opposite/ hypotenuse
= AC/BC
Cos0= adjacent/ hypotenuse
= AB/ BC
Tan0= opposite/adjacent
= AC/ AB
cosec 0= 1/(sin0)
= hypotenuse/ opposite
= BC/ AC
Sec 0= 1/(Cos0)
= hypotenuse/ adjacent
BC/ AB
Cot0= 1/(tan0)
= Adjacent/opposite
= AB/ AC
Secant, cosecant and cotangent are the three functions which are derived from sine, cos, and tan, which are the primary functions. The reciprocal of sine, cos, and tan are cosecant, secant, and cotangent respectively. They are also written as cosec, sec and cot functions.
What are the 6 trigonometric functions to keep in mind while learning this chapter and what are the uses of these functions?
They are Sine, Cosine, Tangent, Secant, Cosecant and Cotangent. These trigonometric functions are used to find the unknown angle or side of a right-angled triangle.
Difference in values of trigonometric functions in different quadrants.
Values of trigonometric functions at angles.
Periodic functions.
Even and odd functions
Once the students understand these, it becomes very easy for them to apply in practical life. Questions based on trigonometry help increase student's thinking abilities and it helps them in the future because it is going to be used in a lot of places. They will need to remember these functions to solve more complex problems.
2. What are the basic trigonometric formulas and their reciprocal identities?
Formulas for Angle θ
Reciprocal Identities
sin θ = Opposite Side / Hypotenuse
sin θ = 1/ cosec θ
cos θ = Adjacent Side / Hypotenuse
cos θ = 1/ sec θ
tan θ = Opposite Side / Adjacent
tan θ = 1/ cot θ
cot θ = Adjacent Side / Opposite
cot θ = 1/ tan θ
sec θ = Hypotenuse / Adjacent Side
sec θ = 1/ cos θ
cosec θ = Hypotenuse / Opposite
cosec θ = 1/ sin θ
3. What are the ratios of all the basic Trigonometric functions and their reciprocals?
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Solution: Given that, sin A = 9/41 …………. (1) Required to find: cos A, tan A By definition, we know that sin A = Perpendicular/ Hypotenuse……………(2) On Comparing eq. (1) and (2), we get Perpendicular side = 9 and Hypotenuse = 41 Let's construct △ABC as shown below, And, here the length of base AB is unRead more
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Are you ready to explore the world of angles and arcs in a circle?
Get ready to discover the differences between minor arcs and major arcs in geometry!
Let's Unravel the Mystery of Angles and Arcs in a Circle!
Angles and arcs in a circle are like pieces of a puzzle that fit together perfectly to create a beautiful picture. In geometry, understanding the properties of angles and arcs is crucial to solving complex problems.
What is the Difference Between Minor Arcs and Major Arcs?
In geometry, a minor arc is an arc that measures less than 180°, while a major arc measures more than 180°. Think of minor arcs as the smaller pieces of the puzzle and major arcs as the larger pieces that complete the picture.
Understanding the difference between minor arcs and major arcs will help you identify and classify arcs in circles accurately. Remember, the measure of an arc determines whether it is a minor or major arc!
Delve Deeper into the World of Geometry!
Geometry is a fascinating branch of mathematics that deals with shapes, sizes, and properties of space. It helps us explore the world around us and understand the relationships between different geometric elements.
By learning about angles and arcs in a circle, you'll gain valuable insights into how geometric concepts work together to form intricate patterns. Whether you're solving puzzles or designing structures, geometry plays a vital role in shaping our world.
So, are you ready to embark on a geometric adventure and uncover the secrets of angles and arcs in a circle?
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8 ... given point A a straight line AL has been drawn equal to the given straight line BC . Which was to be done ... angle BAC equal to the included angle EDF . Then shall the base BC be equal to the base EF ; and the triangle ABC to the ...
Page 13 Euclides Robert Potts. PROPOSITION IX . PROBLEM . To bisect a given rectilineal angle , that is , to divide it into two equal angles . Let BAC be the given rectilineal angle . It is required to bisect it . A D E Take any point D in AB ...
Page 14 ... given straight line , from a given point in the same . Let AB be the given straight line , and C a given point in it ... angle DCF is equal to the angle ECF : ( 1.8 . ) and these two angles are adjacent angles . But when the two ...
Page 15 ... given point C shall be perpendicular to the given straight line AB . Join CF , and CG . And because FH is equal to ... angle FHC is equal to the angle GHC ; ( 1.8 . ) and these are adjacent angles . But when a straight line ...
Page 21 ... given straight lines , but any two whatever of these must be greater than the third . Let A , B , C be the three ... angle equal to a given rectilineal angle . Let AB be the given straight line , and A the given point in it , and
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There is a unique angle \theta between 0^{\circ} and 90^{\circ} such that for nonnegative integers n, the value of \tan \left(2^{n} \theta\right) is positive when n is a multiple of 3 , and negative otherwise. The degree measure of \theta is \frac{p}{q}, where p and q are relatively prime integers. Find p+q.
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What Is a 2D Shape? | Definition & Examples
In the vast realm of geometry, 2D shapes stand as fundamental entities, defining the building blocks of spatial understanding. These flat, two-dimensional figures play a pivotal role in mathematical concepts and have real-world applications, from architecture to art. In this comprehensive exploration, we delve into the characteristics, classifications, and properties of various 2D shapes, shedding light on their significance in both theoretical and practical contexts.
What is 2D shapes?
The essence of 2D shapes lies in their simplicity and flatness. These shapes exist in two dimensions—length and width—without any measurable depth. The boundaries or outlines of these figures are easily distinguishable, making them foundational elements in plane geometry.
2D Shapes Name
1. Circle
A set of all points equidistant from a central point, forming a perfectly round shape.
Circles have a constant radius, diameter, and circumference, with any chord passing through the center being a diameter.
2. Square
A special type of rectangle with all sides of equal length and all angles measuring 90 degrees.
Squares possess symmetry, and each diagonal bisects the opposite angle.
3.Rectangle
A quadrilateral with four right angles, where opposite sides are equal in length and parallel.
Rectangles exhibit symmetry, and their diagonals are of equal length.
4. Triangle
A polygon with three sides and three angles.
Triangles vary in types, such as equilateral, isosceles, and scalene, based on side and angle measurements.
5.Pentagon
A pentagon, derived from Greek, is a polygon with five sides and angles. It exhibits symmetry, with equal sides and interior angles of 108 degrees in a regular form. Beyond geometry, it appears in architecture, art, and nature, showcasing both mathematical elegance and aesthetic appeal. The pentagon, a five-sided wonder, bridges the gap between precision and beauty in the world of shapes.
6.Octagon
An octagon is a polygon with eight sides and eight angles. Its structure is characterized by straight line segments connecting eight vertices, creating an enclosed figure.
2D Shapes Properties
Shape
Properties
Description
Rectangle
Four right angles
A quadrilateral with opposite sides equal and parallel, having four right angles.
Opposite sides equal and parallel
Square
Four equal sides and angles
A special type of rectangle where all sides are equal, forming four right angles.
Triangle
Three sides and angles
A polygon with three sides and three angles, varying in types such as equilateral, isosceles, and scalene.
Circle
Constant radius, diameter, and circumference
A perfectly round shape defined by a set of points equidistant from a central point.
Parallelogram
Opposite sides equal and parallel
A four-sided figure with opposite sides parallel and equal in length.
Trapezoid
One pair of parallel sides
A quadrilateral with one pair of parallel sides and the other two sides not parallel.
Pentagon
Five sides and angles
A polygon with five sides and five angles, contributing to various geometric patterns.
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CMI B.Sc Question Paper
The entrance examination is a test of aptitude for Mathematics at the 12th standard level, featuring both objective questions and problems drawn mostly from the following topics: arithmetic, algebra, geometry, trigonometry, and calculus.
B.Sc Maths & CS Question Paper
Part A:
Part A is worth a total of (4 x10 = 40) points. Points will be given based only on clearly legible nal answers lled in the correct place on page 3. Write all answers for a single question on the designated line and in the order in which they are asked, separated by commas.
Unless specified otherwise, each answer is either a number (rational/ real/ complex) or, where appropriate, one of the phrases \innite"/\does not exist"/\not possible to decide". Write in-teger answers in the usual decimal form. Write non-integer rationals as ratios of two integers.
1. Consider an equilateral triangle ABC with altitude 3 centimeters. A circle is inscribed in this triangle, then another circle is drawn such that it is tangent to the inscribed circle and the sides AB;AC. Infinitely many such circles are drawn; each tangent to the previous circle and the sides AB;AC. The gure shows the construction after 2 steps. Find the sum of the areas of all these circles.
2. List in increasing order all positive integers n <= 40 such that n cannot be written in the form a2 – b2, where a and b are positive integers.
3. How many non-congruent triangles are there with integer lengths a <= b <= c such that a + b + c = 20?
Part B:
Part B is worth a total of 85 points (Question 1 is worth 10 points and the remaining questions are worth 15 points each). Solve these questions in the space provided for each question from page 6. You may solve only part of a question and get partial credit. Clearly explain your entire reasoning. No credit will be given without reasoning.
Answer the following questions
(a) A natural number k is called stable if there exist k distinct natural numbers a1, ……. ;,ak, each ai > 1, such that 1/a1+ ……… +1/ak= 1. Show that if k is stable then k + 1 is also stable. Using this or otherwise, nd all stable numbers.
(b) Let f be a dierentiable function defined on a subset A of the real numbers. Define whenever the above maximum is nite. For the function f(x) = -ln(x), determine the set of points for which f* is defined and find an expression for f*(y) involving only y and constants.
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The range of the line on which the center of the circle may be. These values may be positive, negative, and even infinite. But y must be greater or equal to x.
)
Pushes a circle along a line until it no longer intersects with any edges.
Imagine that you have a balloon at the point point + direction*bias. The balloon may only move along the line from point + direction*inputRange.x to point + direction*inputRange.y. The balloon starts uninflated, but now imagine that you start inflating it. The balloon will inflate like a circle until it intersects with an edge and then when we continue inflating it, we will necessarily have to move the balloon's center since the edge is in the way. We continue like like until we cannot inflate the balloon any further, or we reach the maximum radius. The return value contains the range of the balloon's center along the line and the radius of the balloon.
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Lesson video
Today, I'll be teaching you about combining translations and reflections.
So first make sure you have a pen and paper.
If you have a pencil and rubber and ruler or straight edge, that could be helpful as well.
So pause and make sure you're ready to begin.
Okay.
Remember, you can pause the video whenever you like.
Do pause when I ask you to, but also whenever you need to.
And rewind and listen to bits again if you don't quite understand.
That can sometimes help.
Okay, let's see what the trial this is.
So before I try this, I want you to describe the transformation from A to D, D to C, and C to A.
And C to A is probably the trickiest one.
It might help if you had tracing paper, but if not, I feel like maybe you can still figure a bit out.
And then, use the information that's been given to us, that A to B is a translation by four negative nine.
Can you find a missing coordinate of all of the rectangles? Okay? So this is quite a challenge, but I think it would be really useful to kind of get used to the position grid, different coordinates, and a bit of working.
So do spend five to 10 minutes on this.
Have a go.
Okay.
So from A to D, what did you get? Good.
It was a translation.
Did you write in vector notation? Excellent.
So I would go from negative six three co-ordinate, to four one.
So, how far do you have to go to the right? Excellent.
And then how far down? Good.
So it should be 10 negative two.
What about D to C? Good.
It was reflection, in the line, in the x axis, yes.
So you could do x axis or y equals zero.
They're the same thing.
How about C to A? That was a tricky one.
Rotation, yeah.
Rotations are hard to spot the centre.
What was the centre of rotation? Excellent.
It was negative one one, and it was a 180 degrees.
If you got rotation, how many degrees, then fantastic.
The centre was negative one one.
Excellent, if you've got that as well.
So then all the coordinates, so how did you work out the coordinates first? Good.
So what I would do is see, actually, what is the length of the rectangle, and I will use this, What's this length? So that's the difference in the y values.
So that distance is four.
So, it's going to be ending in seven.
I know that it's on the same line as this other one.
And the exact value is going to be the same.
And then here you can work out this one, because it is negative nine in the x value.
And the y value is going to be the same as this y value.
And then kind of using the reason that we know now that this side is four, and we know that this side is three.
We can then work out all the other ones as well.
That's what you should get.
Pause the video and check your answers.
Okay.
So for the connect, what I want you to do to describe the transformation or combination of transformations between each pair of triangles.
So from S to A, S to T, S to B, B to A, B to T, B to S like that.
Do spend five to six minutes to see how many you can get in that time period.
Okay.
Just gave myself some room to write.
So there are lots you could get.
So let's say from, let's do the axis.
So from S to A, what did you get? It is a reflection.
And, below this line, so imagine points in the line, so we have four one, four two, four three, x equals four.
Good.
What about S to B? Excellent, it's a translation.
So from this to this corresponding point is one two, three downwards, so a translation in the vector zero negative three.
One reason to remember it, it is negative because it's going down.
And then S to T.
No, it's not a rotation.
So some people might have thought rotation about this point, but then actually S would end up looking like, like that.
Like with the right angle up here.
So that is so bad.
Wait a second, sorry, Mrs Buckmire how messy are you? Right.
It would look like, from this point to this point, to this point right here, it would look like that if it's rotation.
So that's not correct.
So this one, we can use a combination.
So what combination could we use? Excellent.
So could go to S A, so it could go S to A, and then A to T.
Or even S to B, and then B to T.
So in that case, it would be a reflection, in x equals four then, I don't know why I used capital it should just be small.
But then a translation of zero negative three.
Okay.
So hold that one in your head from S to T, and let's see what Anthony has to say.
Okay.
Anthony says the order you reflect and translate doesn't matter.
Okay.
So keep what we just did in mind, where we had reflection x equals four, and then translation in, what was it again? Good, zero negative three.
So this was to get from S to T.
Explain why the student is correct in the case of mapping S to T.
Is this statement true for all translations? So in our case would be reflection then translation.
If you did translation first, then reflection, would you still get to T so it says we do, but do you trust them? And it this statement true for all translations, okay.
Just pause and have a think.
Okay.
So, first if we did the translation here first, so we went down three first, so one, two, three, ah, we'd get to B.
And then we did the reflection.
We would end up at T, so actually, if we swap these round, we do, it does work out the order doesn't matter, but will that always be the case? Hmm.
Did you come up with some examples maybe in some of your own questions where it didn't work, maybe where you reflected first, and then you translated.
Interesting.
We'll look more into that later.
Okay.
So for your independent tasks, I do actually want you to look more into it.
Now I'll give you an example, and this is Yasmine.
So Yasmine is a student exploring the effect of combining a translation and a reflection.
And actually she's used a diagonal line reflection here.
So just this line is her line of reflection.
And what we can see is that, first you reflect in the dotted line, and then translate by the vector of negative two two.
So we can see that, for this one.
So she did from A to B one first, so the translation first, and this one, she did translation first.
And then she reflected it and she got to C.
So let me explain.
Where's my.
Yeah, so here, she did A to B one, and that was by a translation.
And then from B one reflected it, so we can see it's one, two, three away, so one, two, three.
So that's how we got to C.
So A to B one to C one so then it was a reflection.
So then you'll just reflect it first, so a reflected one, two, three, so reflected one, two, three.
So I got to here, and then she did the translation of negative two two.
So negative two two.
And she ended up at C.
So, A to B two, to C, was a reflection first, and then a translation, okay? So what I'm going to do is I'm going to give you some reflection, a reflection line, a line reflection, and a translation.
I want you to do it in different orders, okay? Hopefully that makes sense.
Okay.
So here, use the line of symmetry shown to compare the effect of reflecting then translating, and then of translating then reflecting, for each of the vectors, one zero, one two, and one one, with the same line of symmetry.
Okay.
So, here it is, you can do it on separate diagrams. If you like, you can just do it all on the same diagram.
So this is for one zero.
This is on a worksheet, for worksheets if you want to use it off the video, this is for one one, this is for one two, And that is you've finished.
So do make sure you pause there, look at the worksheet and do how I've done it.
Okay.
So question one.
So reflecting and then translating, so let's see which answer's which.
So here, this green one, let me do it in green so it's the same.
This is it having been translated right now.
So we're translating it to one zero.
So that's one to the line, makes sense.
And then it's being reflected, so it ends up here, right.
So, the red one, this is the reflection, ends up here.
And then when it is, so this is the final one, let me label the final one A there, to make it easier.
And then when it's being translated, it translates to here, so it's not the same.
So here, the order did matter.
Hmm.
Interesting.
Okay.
What about here? So here, for one one, when it was done first, which is A, you end up here, and then the final one, ended up being here when it was reflected.
So that was the final.
So, and then what about the other one? So it was reflected first, it ends up here, and then it was moved one one.
So that's one right, one up.
Oh, it ends up in the same place.
So it did work for this one.
Didn't work for the first one, the order didn't match up, in the second one, the order didn't matter.
That's strange.
I wonder why.
Hmm, do you notice anything between one zero and one one?> I wonder.
Compared to the line of symmetry maybe.
Okay, I just want you to show you this one.
So if it's reflected first, it'll be there, and then it's translated one two, that's one to the right and two up.
So it moves to here.
So that'll be your final one.
Or if it translated first, I moved one and then two up.
So it's like up here off the grid, and then it is reflected.
So it'll be here off the grid, but the other one was not off the grid.
So it doesn't work.
One two didn't work either.
So only one one worked, with that diagonal line.
Okay.
So the shape undergoes translation or reflection.
When do your following combinations have the same effect? So we've seen lots of cases where it has worked and somewhere it hasn't worked.
So why is that? What I would like you to do is look back at those examples and maybe even create some of your own examples, and just think, okay, what do I notice? What's the relationship here? So for, do they have the same relationship? Does it have the same effect when we do a translation then the reflection, or the reflection then the translation.
Just have a look through our notes or look back at certain examples, you could scroll through.
Pause the video and just think about it.
Okay.
So, hopefully you had a chance to have a look at those different examples.
So I'm going to show you this example.
So this was one that did work, and this was from independent task, and it was when we had that diagonal one, the one we had one one, and this was also one that worked.
So here it worked when we had the translation between zero negative three and this vertical line, which is x equals four.
Someone told me once, and when I did this, but maybe it has something to do with the line of reflection and whether the vector's parallel.
What do you mean by parallel? So parallel, they're lines that never meet, but let's say so here's zero negative three, if I did that as an arrow, that would just be a downward arrow.
From here to here.
What would one one look like? So we'd go one across, and one up.
So it'd be from here to here, to look like that.
So what do we notice about the arrows, and the lines of symmetry considering what I just said to be fair? There are parallel.
So parallel means they never meet.
They're parallel to each other.
I wonder if this is always the case.
Maybe you could have another go at this activity, using two two, cause that would be parallel, going two across then two up.
Or maybe this activity using zero two, and then x four swapping those rounds.
So pause, if you want to check out that hypothesis, that actually, if the vector is parallel to the line of reflection, and the order doesn't matter.
Okay.
Did you have a go? It does work out, doesn't it? Interesting.
So if I was going to actually write it as an official mathematical statement, what I would say is, the combined transformation equivalent to a translation and a reflection is commutative, when the translation vector is parallel to the line of reflection, so many words, but do you know what? You understand and know all of them.
Commutative, you've learned that before.
So A plus B equals B plus A? Is there any operation where the order doesn't matter? And then parallel must be in the same direction so it wouldn't meet.
If it was a line, it wouldn't meet.
So parallel to the line of reflection, which, you know, the translation vector.
Wow.
You've understood that.
Fantastic.
Now, even if you didn't, I think you've worked really hard today, and it's really good.
Even if you just had a go at translating and reflecting, you just practise that.
That's really, really great.
So really, really well done.
Hopefully you got to check over your work.
And, I would love it for you to do the exit quiz because that will help with your learning.
And I'd just like to say thank you for participating in the lesson today.
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Distance Formula: Finding the Distance Between Two Points
The distance formula is an algebraic expression that gives the shortest distance between two points in a two-dimensional space. Dream01/Shutterstock/HowStuffWorks
You're sitting in math class trying to survive your latest pop quiz. Sweat trickles down your forehead as you read the prompt: "Find the distance between these points."
The distance formula you're looking for is fairly straightforward and has ties to one of the most useful and famous concepts in all of mathematics: the Pythagorean theorem.
What Is the Distance Formula?
The distance formula is an algebraic equation used to find the length of a line segment between two points on a graph, called the Cartesian coordinate system (also known as the point coordinate plane).
This two-dimensional plane is defined by two perpendicular axes (usually labeled the x-axis and the y-axis) that intersect at a central point called the origin. Here's how it's expressed:
In a two-dimensional space with two points P (x₁, y₁) and Q(x₂, y₂), the distance (d) between these two points is given by the formula: d = √ (x₂ - x₁)² + (y₂ - y₁)²
In a three-dimensional space with two points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂), the distance (d) between these two points is given by the formula: d = √ (x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²
Next up, we'll take a closer look at the point coordinate plane, which can help you find exact spots by their horizontal and vertical positions, essential for everything from math problems to GPS navigation.
Understanding the Point Coordinate Plane
When most people hear the word "graph," they're picturing a chart with two lines — one vertical, one horizontal — that intersect each other at a right angle.
The vertical line is called the y-axis, and its horizontal counterpart is the x-axis. Both lines work together to tell a story with data.
If you want to make sense of where one point rests on your graph, measure where it falls along the two dimensions (the x-axis and the y-axis). These are known as the point's coordinates.
You need to find the coordinates for the first point and the second point before you can calculate the distance between them. You'll use the distance formula to measure the straight line segment connecting the two points.
Now let's explore the blissful relationship between the Pythagorean theorem and the distance formula.
The Pythagorean Theorem and the Distance Formula
The Pythagorean theorem was named for the Greek philosopher Pythagoras, but over a millennium before his birth, the ancient Babylonians already understood the geometric principle that is now associated with his name.
In essence, the Pythagorean theorem tells us how to find the longest side of a triangle when we know the lengths of the other two sides, and the distance formula uses this idea to measure how far apart two points are on a graph by treating the points as if they're at the corners of a right triangle.
For those in need of a quick refresher, the Pythagorean theorem says: The area of the square built upon the hypotenuse of a right triangle is equal to the sum of the areas of the squares upon the remaining sides.
There are a few key points to understand here. A right triangle, or right-angled triangle, has one angle measuring 90 degrees, known as a right angle. The longest side of this triangle is termed the hypotenuse, which is located opposite the right angle.
As we all know, a triangle may have three sides, but a square has four. So, imagine taking the hypotenuse of a right triangle and turning it into one of the four lines of a brand-new square. Then, do the same thing to the other two sides in the original triangle. You'll end up with three individual squares.
According to the Pythagorean theorem, the square formed by the hypotenuse has an area equal to the sum of the areas of the squares formed by the other two sides. If the hypotenuse is labeled "c," and the other two sides are labeled "a" and "b," then we could express that idea like so:
The Pythagorean theorem says a2 + b2 = c2. The distance formula is derived by using the theorem. grebeshkovmaxim/Shutterstock
How to Find the Distance Between Two Points
The first point and second points on your graph will each have an x coordinate and a y coordinate. You can calculate the shortest distance between these two points by using the Euclidean distance formula, which is a Pythagorean theorem-related algebraic expression.
D = √(x₂ - x₁) ² + (y₂ - y₁)²
Here, D stands for "distance." As for x₂ and x₁, they refer to the x coordinates of Point 2 and Point 1, respectively. Same goes for y₂ and y₁, except those are the two y coordinates.
So, to calculate the distance, our first step is to subtract x₁ from x₂. Then we have to multiply the resulting number by itself (or, in other words, "square" that number).
After that, we must subtract y₁ from y₂ and then square the answer we get from doing so. This will leave us with two numbers we must add together.
Then, finally, take that number and find its square root. And that square root, ladies and gentlemen, is our distance.
Distance Formula Example
OK, so let's say Point A has an x coordinate of 2 and a y coordinate of 5 (2,5). Let us also assume that Point B's got an x coordinate of 9 and a y coordinate of 13 (9,13). Plug those values into the handy-dandy formula and you get this:
D = √(9-2)² + (13-5)²
What's 9 minus 2? Easy, 7. And 13 minus 5 is 8, of course.
Now we're left with this:
D = √7² + 8²
If you "square" 7 — as in, multiply the number by itself — you end up with 49. As for 8 squared that works out to 64. Let's plug those values into the equation, eh?
D = √49 + 64
Now we're cooking. Add 49 and 64, and you get 113.
D = √113
What's the square root of 113? The answer is 10.63, so therefore:
D = 10.63
Go forth and ace your next pop quiz!
This article was updated in conjunction with AI technology, then fact-checked and edited by a HowStuffWorks editor
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Description: The students will be shown a video on software. They will then be asked questions as it relate to Software and its purpose as shown in the video as well as what they already know. The students will be required to brainstorm and provide responses.
The students will then be shown information about the System and Application Software. The students will read through the information. The students will be placed in groups of ten and each group will be given two questions that they need to answer using the information given. They will use the method Think-pair share technique at this time. After reading through the information, each group will provide the answers to the questions. In addition, the students should listen critically to and questions other students explanations of the major concepts and principles taught.
The teacher will ascertain students comprehension of the topic and justify or clarify any misconception that a student may have in this topic
Description: 1. Issue each group with the compass point for them to discuss among themselves and identify and record what they need to know about angles in a polygon as well as what excites them, what worries them and what current stance or ideas they have on the topic.
2. Each group will be issued with a polygon. Two groups will be given pentagons, two will be given hexagon, two groups will be given heptagon and one group will be given an octagon. Each group will try to give the name of their polygon and a reason for its name. Teacher will correct and clarify as necessary.
3. Students will be asked to suggest ways that we can figure out the sum of the interior angles of their polygon without the use of a protractor. After taking suggestions, groups will then be instructed to see how many triangles can be formed by connecting one vertex to each of the other vertices using straight line segments. These line segments are called diagonals.
4. Each group will share and the results placed in a table. The sum of interior angles column will be left blank.
Name of polygon Number of sides Number of triangles Sum of Interior angles. Quadrilateral Pentagon Hexagon Heptagon Octagon
5. From the table students will see if they can see any pattern in the number of sides on a polygon and the number of triangles formed. Students should note that in each case the number of triangles is two less than the number of sides.
6. Teacher will ask students if this information about the number of triangles can be useful in finding out the sum of the interior angles in the polygon and if yes, how?
7. Teacher will explain that since interior angles in the triangle add up to 180 degrees, the number of triangles can be multiplied by 180 to get the sum of interior angles in the given polygon.
8. The stop sign and the school road signs will be displayed for students to state the name of the polygon. Teacher will ask if by their estimation each angle in the polygons are equal. From this, students will see that some polygons are regular while others are irregular. Regular polygons have all their angles equal while irregular ones don't.
Description: Engage: Students will select a paper from a black purse bag. The paper will have a number and information on climate change whether a supporting or opposing viewpoint. 24 students will select a number from the bag. 12 for and 12 against climate change. The remaining ten students will be judges. The students will be given the criteria for the debate. Explore: Students will read the information on their cards and then will decide if they are supporting or opposing. The teacher would have the designated areas for support and opposition setup while the students read their cards. The two groups will then be further divided into 8 groups. Four groups supporting climate change and four groups in opposition. Groups will have 3 members: Group 1: 1, 2 and 3 Group 2: 4, 5 and 6 Group 3: 7, 8 and 9 Group 4: 10, 11 and 12 The group will have 20 minutes to discuss their points and group members can exchange or swap their information as they see fit. Explain: The students will then debate their points in front of the panel of judges (the remaining students). There will be two judges for each debate along with the teacher. Each student will present their case and some of the unoccupied students will ask students on the panel some questions. Elaborate: The unoccupied students in the audience will ask the students on the panel questions based on what was presented. Asking them to provide further clarification to certain points.
Description: Students will participate in a game of Chinese telephone. Teacher or a student will whisper a message to another student and the message is whispered to each student until all students get the message. The last student to receive the message will say it aloud. This message will be compared with the original message. The students will be led into a guided discussion for them to provide definitions for the concepts based on participation in the game; the discussion will also center around why the message changed or did not change. This will be noted on the board. Students will be placed in groups of six (6). Students will use their devices to conduct research on the forms of communication on their devices. The stronger students will work with the slower ones. Students will choose representatives from their groups to share the information they found on the forms of communication. Stronger students will be chosen to construct a concept map that will show the different forms of communication and examples of each. Teacher will assist students in constructing this. They will be shown pictures of the different forms of communication.
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Sides of a Triangle Calculator: An In-Depth Guide
In geometry, a triangle is a polygon with three edges and three vertices. It is one of the basic shapes in geometry and is used in various applications, including architecture, engineering, and design. The sides of a triangle play a crucial role in determining its properties and characteristics. This informative article delves into the world of triangle side calculators, exploring their functionality, benefits, and practical applications.
Calculating the sides of a triangle can be a tedious and time-consuming task, especially when dealing with complex triangles. Triangle side calculators offer a convenient and efficient solution to this problem. These online tools are designed to swiftly compute the lengths of triangle sides based on the given parameters, such as angles or other side lengths. They employ sophisticated algorithms and mathematical formulas to provide accurate and reliable results in a matter of seconds.
Moving forward, we will delve deeper into the realm of triangle side calculators, exploring their features, benefits, and real-world applications. We will uncover the mathematical principles behind these calculators and demonstrate how they can be harnessed to solve various geometric problems.
Triangle side calculators have revolutionized the way we solve triangle problems by providing effortless calculations that simplify even the most complex computations.
Eliminates Tedious Calculations:
Triangle side calculators eliminate the need for manual calculations, which can be time-consuming and prone to errors. By automating the process, these tools save valuable time and reduce the risk of mistakes.
Handles Complex Formulas:
Triangle side calculators are equipped to handle complex formulas and algorithms, allowing users to solve problems that would otherwise require advanced mathematical knowledge. This makes them accessible to individuals of all skill levels.
Covers Diverse Triangle Types:
These calculators can handle various types of triangles, including right triangles, isosceles triangles, equilateral triangles, and scalene triangles. They automatically apply the appropriate formulas based on the given parameters.
Provides Step-by-Step Solutions:
Many triangle side calculators offer step-by-step solutions, allowing users to follow the calculation process and understand the underlying principles. This feature is particularly helpful for students and learners who want to reinforce their understanding of triangle geometry.
The effortless calculations provided by triangle side calculators make them invaluable tools for anyone working with triangles, from students and educators to engineers and architects. They streamline the problem-solving process, enhance accuracy, and promote a deeper understanding of triangle geometry.
Time-Saving Tool: Delivers Results in Seconds, Boosting Efficiency
Triangle side calculators are renowned for their lightning-fast calculations, delivering results in mere seconds. This remarkable speed offers significant advantages in terms of efficiency and productivity.
Rapid Problem-Solving:
With triangle side calculators, users can solve complex triangle problems in a matter of seconds. This rapid problem-solving capability is particularly beneficial in time-sensitive situations or when dealing with a large number of triangles.
Boosts Productivity:
By eliminating the need for manual calculations, triangle side calculators significantly boost productivity. This allows users to focus on other tasks or solve more problems in the same amount of time.
Enhances Learning Efficiency:
For students and learners, triangle side calculators can enhance learning efficiency by providing quick and accurate solutions. This allows them to spend less time on calculations and more time understanding the underlying concepts and principles of triangle geometry.
Facilitates Real-Time Applications:
In real-time applications, such as engineering or architecture, triangle side calculators enable professionals to obtain results instantaneously. This facilitates rapid decision-making and ensures the accuracy of calculations, which is crucial in these fields.
The time-saving capabilities of triangle side calculators make them indispensable tools for anyone who works with triangles, from students and educators to professionals in various fields. By delivering results in seconds, these calculators streamline the problem-solving process, enhance productivity, and promote efficient learning and decision-making.
Accuracy Guaranteed: Employs Advanced Algorithms for Precise Outcomes
Triangle side calculators employ sophisticated algorithms and mathematical formulas to deliver precise and accurate results. This unwavering accuracy is a hallmark of these tools and makes them invaluable in various applications.
Reliable Calculations:
Triangle side calculators leverage advanced algorithms to ensure the reliability of their calculations. This means users can trust the results provided by these tools, even when dealing with complex or intricate triangle problems.
Eliminates Human Error:
By automating the calculation process, triangle side calculators eliminate the possibility of human error. This is particularly important in situations where accuracy is paramount, such as in engineering, architecture, or scientific research.
Consistent Results:
Triangle side calculators provide consistent results, regardless of the user or the number of times the calculation is performed. This consistency ensures that users can rely on the accuracy of the results, even when dealing with large datasets or repetitive calculations.
Adherence to Mathematical Principles:
Triangle side calculators adhere strictly to mathematical principles and formulas. This ensures that the results obtained are mathematically sound and accurate, providing users with confidence in the validity of their calculations.
The accuracy guaranteed by triangle side calculators is a testament to their reliability and dependability. By employing advanced algorithms and adhering to mathematical principles, these tools deliver precise and trustworthy results, making them indispensable for anyone working with triangles, from students and educators to professionals in various fields.
Diverse Parameters: Accepts Various Inputs Like Angles, Side Lengths, and Ratios
Triangle side calculators are designed to handle a wide range of inputs, including angles, side lengths, and ratios. This versatility makes them suitable for solving a variety of triangle problems.
Angles:
Triangle side calculators can accept angle measurements in degrees, radians, or grads. This allows users to solve problems involving angles of any величина, regardless of the units used.
Side Lengths:
Triangle side calculators can also accept side lengths as inputs. This enables users to find the lengths of unknown sides based on the given information. The calculators support various units of measurement, such as inches, centimeters, meters, and feet, providing flexibility in handling different scenarios.
Ratios:
In some cases, triangle side calculators allow users to input ratios between side lengths. This is particularly useful when working with similar triangles, where the ratios of corresponding sides are equal. By providing ratios, users can solve problems involving proportional triangles.
The diverse input parameters accepted by triangle side calculators make them adaptable to a wide range of triangle problems. Whether users have information about angles, side lengths, or ratios, these calculators can handle the inputs and deliver accurate results.
The versatility of triangle side calculators extends beyond the types of inputs they accept. These calculators also support a variety of triangle types, including right triangles, isosceles triangles, equilateral triangles, and scalene triangles. This allows users to solve problems involving any type of triangle, making triangle side calculators truly comprehensive tools for triangle-related calculations.
Triangle side calculators employ a collection of formulas to compute side lengths based on the given input parameters. These formulas are carefully selected and applied to ensure accurate and reliable results.
Pythagorean Theorem:
For right triangles, triangle side calculators utilize the Pythagorean theorem, which states that the square of the hypotenuse is equal to the sum of the squares of the other two sides. This formula is particularly useful for finding the length of the hypotenuse or the other sides of a right triangle.
Law of Sines:
The law of sines is applied in triangles where two angles and one side are known. This formula establishes a relationship between the ratios of side lengths and the sines of their opposite angles. It allows users to find the lengths of unknown sides or angles in a triangle.
Law of Cosines:
The law of cosines is used when two sides and the included angle of a triangle are known. This formula involves the relationship between the squares of the sides and the cosine of the included angle. It is particularly useful for finding the length of the third side in a triangle.
Area Formula:
Triangle side calculators also employ the area formula to find the area of a triangle when two sides and the included angle are known. This formula uses the product of half the product of the two sides and the sine of the included angle to calculate the area.
Triangle side calculators seamlessly select and apply the appropriate formula based on the input parameters provided. This ensures that the results are accurate and consistent, regardless of the type of triangle or the information available.
Step-by-Step Solutions: Provides Detailed Breakdowns of Calculations
Many triangle side calculators offer a valuable feature called step-by-step solutions. This feature provides a detailed breakdown of the calculations, allowing users to follow the process and understand how the results are obtained.
Benefits of Step-by-Step Solutions:
Enhanced Understanding:
Step-by-step solutions help users develop a deeper understanding of the mathematical principles and formulas used in triangle calculations. By following the step-by-step process, users can see how the input parameters are manipulated and transformed to arrive at the final result.
Error Identification:
The detailed breakdown of calculations allows users to identify any potential errors or misunderstandings in their approach. By examining each step, users can pinpoint where an error may have occurred and make necessary corrections, leading to more accurate results.
Learning Tool:
Triangle side calculators with step-by-step solutions serve as excellent learning tools for students and individuals who want to improve their understanding of triangle geometry. By studying the step-by-step process, users can reinforce their knowledge of mathematical concepts and apply them effectively in solving triangle problems.
Verification of Results:
The step-by-step solutions provide a means for users to verify the accuracy of their results. By checking each step and ensuring that the calculations are performed correctly, users can gain confidence in the validity of their answers.
Conclusion:
The step-by-step solutions offered by triangle side calculators are a valuable asset for users who seek to understand the underlying principles of triangle calculations and enhance their problem-solving skills. These detailed breakdowns promote learning, error identification, result verification, and overall proficiency in triangle geometry.
In addition to the benefits mentioned above, step-by-step solutions can also be helpful for users who need to document their work or share their calculations with others. By providing a clear and organized breakdown of the process, step-by-step solutions facilitate communication and collaboration among individuals working on triangle-related problems.
Interactive Interface: User-Friendly Design for Ease of Use
Triangle side calculators are designed with intuitive layouts and clear instructions, making them easy to navigate and use. Users can quickly identify the input fields, select the appropriate formulas, and obtain results without any hassle.
Interactive Inputs:
Many triangle side calculators feature interactive input fields that provide real-time feedback. As users enter values, the calculator automatically updates the results, allowing them to explore different scenarios and make adjustments on the fly.
Visual Aids:
Some triangle side calculators incorporate visual aids, such as diagrams and graphs, to help users visualize the triangle and understand the relationships between its sides and angles. This visual representation can be particularly helpful for learners and individuals who prefer a more graphical approach.
Error Handling:
Triangle side calculators often include error handling mechanisms to alert users to incorrect or missing inputs. These calculators provide clear error messages, guiding users to identify and rectify any issues, ensuring accurate and reliable results.
The interactive interface of triangle side calculators makes them accessible to users of all skill levels. Whether you are a student learning about triangle geometry or a professional working on complex triangle-related problems, these calculators provide a user-friendly environment that facilitates efficient and effective problem-solving.
Wide Applications: Applicable in Geometry, Engineering, Architecture, and More
In geometry, triangle side calculators are indispensable tools for solving triangle-related problems. Students and educators use them to explore concepts such as the Pythagorean theorem, the laws of sines and cosines, and area calculations.
Engineering:
In engineering, triangle side calculators are used in structural analysis, mechanical design, and fluid dynamics. Engineers rely on these calculators to determine forces, moments, and stresses in triangular structures and components.
Architecture:
In architecture, triangle side calculators are used in roof design, truss analysis, and space planning. Architects use them to calculate angles, lengths, and areas of triangular elements in buildings and structures.
Surveying:
In surveying, triangle side calculators are used to determine distances, angles, and elevations. Surveyors use them to create maps, measure property boundaries, and conduct topographical surveys.
The versatility of triangle side calculators extends to other fields as well. They are used in carpentry, woodworking, graphic design, and even in recreational activities like puzzles and games. Their ability to handle a wide range of triangle problems makes them valuable tools across diverse disciplines.
FAQ
Welcome to the Frequently Asked Questions (FAQ) section for triangle side calculators!
Question 1: What is a triangle side calculator?
Answer: A triangle side calculator is an online tool that helps users find the lengths of the sides of a triangle based on the given information. These calculators employ advanced algorithms and mathematical formulas to provide accurate and reliable results.
Question 2: What types of triangles can be solved using a triangle side calculator?
Answer: Triangle side calculators can handle various types of triangles, including right triangles, isosceles triangles, equilateral triangles, and scalene triangles. They automatically apply the appropriate formulas based on the input parameters.
Question 3: What information do I need to use a triangle side calculator?
Answer: Depending on the calculator and the type of triangle, you may need to provide information such as the lengths of two sides, the measure of an angle, or the ratio of two sides. Some calculators also allow you to input side lengths and angles simultaneously.
Question 4: Are triangle side calculators accurate?
Answer: Yes, triangle side calculators are generally accurate when used correctly. They employ sophisticated algorithms and adhere to mathematical principles to ensure precise and reliable results.
Question 5: Can I use a triangle side calculator on my phone or tablet?
Answer: Many triangle side calculators are web-based and can be accessed on various devices, including phones, tablets, and computers, as long as you have an internet connection.
Question 6: Are triangle side calculators free to use?
Answer: Yes, most triangle side calculators are free online tools. You can access and use them without any subscription or registration fees.
Question 7: Can triangle side calculators help me learn about triangle geometry?
Answer: Yes, triangle side calculators can be valuable learning tools. They can help you understand the relationships between the sides and angles of a triangle and reinforce your knowledge of triangle geometry concepts.
Closing:
We hope this FAQ section has answered your questions about triangle side calculators. If you have any further inquiries, feel free to search for more information online or consult a mathematics educator or professional.
Now that you know more about triangle side calculators, explore our next section for helpful tips and tricks to make the most of these tools and enhance your triangle-solving skills.
Tips
Here are some helpful tips to make the most of triangle side calculators and enhance your triangle-solving skills:
Tip 1: Choose the Right Calculator:
Select a triangle side calculator that suits your needs and skill level. Some calculators offer basic functions, while others provide advanced features and options. Consider the types of triangles you typically work with and choose a calculator that can handle those specific scenarios.
Tip 2: Understand the Input Parameters:
Familiarize yourself with the input parameters required by the triangle side calculator. Ensure that you have the necessary information, such as side lengths, angles, or ratios, before using the calculator.
Tip 3: Check for Accuracy:
While triangle side calculators are generally accurate, it's always a good practice to double-check the results, especially when dealing with complex problems. Use alternative methods or a different calculator to verify your answers.
Tip 4: Explore Step-by-Step Solutions:
If the triangle side calculator provides step-by-step solutions, take advantage of this feature. It can help you understand the underlying mathematical principles and formulas used in the calculations. This can be particularly beneficial for learning purposes or when troubleshooting errors.
Closing:
By following these tips, you can effectively utilize triangle side calculators to solve triangle problems with ease and accuracy. Remember, practice makes perfect. The more you use these calculators and explore their features, the more proficient you will become in solving triangle-related problems.
With the knowledge gained from this article, you are now equipped to harness the power of triangle side calculators and tackle triangle problems confidently. Head to the conclusion section for a brief summary and final thoughts on these invaluable tools.
With triangle side calculators at our fingertips, we can confidently tackle triangle-related problems, whether for academic pursuits, professional endeavors, or personal curiosity. These tools empower us to explore the fascinating world of triangle geometry, understand the relationships between sides and angles, and solve complex problems with ease. Embrace the power of triangle side calculators and unlock your full potential in solving triangle problems.
Remember, the key to mastering triangle side calculators lies in practice and exploration. The more you use these tools and delve into their features, the more proficient you will become in solving triangle problems. So, continue your journey of discovery, sharpen your skills, and enjoy the satisfaction of solving triangle problems with accuracy and efficiency.
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A tilted right circular cone with apex angle $2 \theta_c = 45^\circ$ (where $\theta_c$ is the semi-vertical angle), has its apex at the point $A(a_x, a_y, a_z)$ with $a_x, a_y, a_z \gt 0$, such that its intersection with the $xy$ plane is given by
$$ 5 x^2 - 4 xy + 9 y^2 = 64 $$
Find the coordinates of the apex $A$.
My attempt:
First I identified the major and minor axes of the ellipse of intersection, their direction, and the eccentricity. From the eccentricity and the known semi-vertical angle, I calculated the angle between the normal to the $xy$ plane and the axis of the cone. Then used the angle bisector theorem to deduce the location of the point where the axis intersects the $xy$ plane and the distance between the apex and that intersection point. From there, I computed the coordinates of the apex.
$\begingroup$From Cone on wikipedia $$((x-a_x,y-a_y,z-a_z)\cdot(d_x,d_y,d_z))^2-(d_x,d_y,d_z)\cdot(d_x,d_y,d_z)(x-a_x,y-a_y,z-a_z)\cdot(x-a_x,y-a_y,z-a_z))/2=0$$ where $(d_x,d_y,d_z)$ is a vector parallel to the axis.$\endgroup$
Remember $\frac{\sqrt{2+\sqrt2}}{2}=\cos(\frac{\pi}{8})=\frac{n_x x+ n_y y+n_z z}{\sqrt{x^2+y^2+z^2}}$ where $\hat{n}=(n_x,n_y,n_z)$ is a unit vector along the axis. We get $$\frac{2+\sqrt2}{4}(x^2+y^2+z^2)=(n_x x+ n_y y+n_z z)^2\tag3$$
Comparing coefficients in $(2)$ and $(3)$, where $k$ is a scaling factor, we get the following system, solving it in maxima CAS, precomputed a grobner basis in M2, all the while remembering that $$n_x^2+n_y^2+n_z^2=1$$
$\begingroup$That's what I did in my solution. I looked at the last equation in your solution (with $z = 0$), and deduced $a_x, a_y, a_z$, which you used in your solution to find the equation of the cone. However, the question doesn't ask for the equation of the cone, just its apex coordinates.$\endgroup$
The eccentricity of the conic section and the angles $\theta_c$ (the semi-vertical angle) and $\phi$ (the angle between the normal to the section plane and the axis of the cone) is given by this important relation
$ e = \dfrac{ \sin(\phi) }{\cos(\theta_c) }$
Using this relation, we can compute $\phi$ readily. It comes to
$ \phi = \sin^{-1} \left( e \cos(\theta_c) \right) \approx 0.77666 $
Now, if we draw a section of the ellipse in the plane $y = \tan(\theta_0) x $, it will be the triangle shown in the diagram below.
where $\psi = \dfrac{\pi}{2} - \phi = 0.79414 $
Applying the law of sines on the triangles on both sides of the angle bisector (with is the axis of the cone, and is of length $z_0$) we get
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NCERT solutions for exercise 11.3 Class 12 Maths chapter 11 move around the topic plane. The questions in NCERT solutions for Class 12 Maths chapter 11 exercise 11.3 are related to exercise 11.3 Class 12 Maths equation of a plane in different conditions, the concept of coplanarity of two lines, the angle between two planes and the exercise 11.3 Class 12 Maths also covers the distance between a point and a plane. One should grasp the concepts well before solving Class 12 Maths chapter 11 exercise 11.3. And to get more idea about steps involved in solving the problems under the topic plane, one can go through the solved example given in the NCERT and then crack the Class 12th Maths chapter 11 exercise 11.3. The below-mentioned exercises are also present in the NCERT Book.
Dimensional geometry 11.1
Three Dimensional geometry 11.2
Three Dimensional geometry miscellaneous exercise
Three Dimensional Geometry Class 12th Chapter 11-Exercise: 11.3
Question:1(a) In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
z = 2
Answer:
Equation of plane Z=2, i.e.
The direction ratio of normal is 0,0,1
Divide equation by 1 from both side
We get,
Hence, direction cosins are 0,0,1.
The distance of the plane from the origin is 2.
Question:1(b) In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
x + y + z = 1
Answer:
Given the equation of thec) In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
2x + 3y - z = 5
Answer:
Given the equation of plane isd) In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
5y + 8 = 0
Answer:
Given the equation of2 Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector .
Answer:
We have given the distance between the plane and origin equal to 7 units and normal to the vector .
So, it is known that the equation of the plane with position vector is given by, the relation,
, where d is the distance of the plane from the origin.
Calculating ;
is the vector equation of the required plane.
Question:3(a) Find the Cartesian equation of the following planes:
Answer:
Given the equation of thebc4(a) In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
2 x + 3y + 4 z - 12 = 0
Answer:
Let the coordinates of the foot of perpendicular P from the origin to the plane be
Given a plane equation ,
Or,
The direction ratios of the normal of the plane are 2, 3b) In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
3y + 4z - 6 = 0
Answer:
Let the coordinates of the foot of perpendicular P from the origin to the plane be
Given a plane equation ,
Or,
The direction ratios of the normal of the plane are 0,3c) In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
x + y + z = 1
Answer:
Let the coordinates of the foot of perpendicular P from the origin to the plane be
Given plane equation .
The direction ratios of the normal of the plane are 1,1 and 1 ..
Question: 4(d) In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
5y + 8 = 0
Answer:
Let the coordinates of the foot of perpendicular P from the origin to the plane be
Given plane equation .
or written as
The direction ratios of the normal of the plane are 0, -5 and 0 .
Question:5(a) Find the vector and cartesian equations of the planes (a) that passes through the point (1, 0, – 2) and the normal to the plane isor
So, this is the required Cartesian equation of the plane.
Question:5(b) Find the vector and cartesian equations of the planes
that passes through the point (1,4, 6) and the normal vector to the plane is .So, this is the required Cartesian equation of the plane.
Question:6(a) Find the equations of the planes that passes through three points.
(1, 1, – 1), (6, 4, – 5), (– 4, – 2, 3)
Answer:
The equation of the plane which passes through the three points is given by;
Determinant method,
Or,
Here, these three points A, B, C are collinear points.
Hence there will be an infinite number of planes possible which passing through the given points.
Question:6(b) Find the equations of the planes that passes through three points.
(1, 1, 0), (1, 2, 1), (– 2, 2, – 1)
Answer:
The equation of the plane which passes through the three points is given by;
Determinant method,
As determinant value is not equal to zero hence there must be a plane that passes through the points A, B, and C.
Finding the equation of the plane through the points,
After substituting the values in the determinant we get,
So, this is the required Cartesian equation of the plane.
Question:7 Find the intercepts cut off by the plane 2x + y – z = 5.
Answer:
Given plane
We have to find the intercepts that this plane would make so,
Making it look like intercept form first:
By dividing both sides of the equation by 5 (as we have to make the R.H.S =1) , we get then,
So, as we know that from the equation of a plane in intercept form, where a,b,c are the intercepts cut off by the plane at x,y, and z-axes respectively.
Therefore after comparison, we get the values of a,b, and c.
.
Hence the intercepts are .
Question:8 Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.
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Is the support vector a position vector and why?
No, the support vector is not a position vector. In machine learning, a support vector is a data point that lies closest to the de...
No, the support vector is not a position vector. In machine learning, a support vector is a data point that lies closest to the decision boundary separating different classes in a classification problem. It is used to define the optimal hyperplane that maximizes the margin between classes. Therefore, a support vector is not a position vector in a geometric sense, but rather a key component in determining the decision boundary in a support vector machine algorithm.
What does the normal vector say in vector representation?
The normal vector in vector representation represents the direction perpendicular to the surface of the object or plane. It is a v...
The normal vector in vector representation represents the direction perpendicular to the surface of the object or plane. It is a vector that is orthogonal to the surface and points outward from the surface. The normal vector is used in various mathematical and physical applications, such as in calculating the direction of force or in determining the orientation of a surface.
Source:AI generated from FAQ.net
Scalar or vector?
Scalar or vector? Scalars are quantities that have only magnitude, such as mass or temperature. Vectors are quantities that have b...
Scalar or vector? Scalars are quantities that have only magnitude, such as mass or temperature. Vectors are quantities that have both magnitude and direction, such as velocity or force.
Source:AI generated from FAQ.net
What is the difference between vector AB and vector BA?
The difference between vector AB and vector BA lies in their direction. Vector AB represents the displacement from point A to poin...
The difference between vector AB and vector BA lies in their direction. Vector AB represents the displacement from point A to point B, while vector BA represents the displacement from point B to point A. Despite having the same magnitude, these vectors have opposite directions, making them distinct entities in terms of orientation Premium position vector and zero vector?
A position vector represents the location of a point in space relative to a reference point or origin, and it has both magnitude a...
A position vector represents the location of a point in space relative to a reference point or origin, and it has both magnitude and direction. On the other hand, a zero vector has a magnitude of zero and represents a point in space that has no displacement from the origin. In other words, a position vector points to a specific location in space, while a zero vector points to the origin and has no physical significance in terms of displacement.
Source:AI generated from FAQ.net
What is the difference between position vector and support vector?
A position vector is a vector that represents the position of a point in space relative to a reference point or origin. It specifi...
A position vector is a vector that represents the position of a point in space relative to a reference point or origin. It specifies the location of a point in terms of its distance and direction from the origin. On the other hand, a support vector is a concept used in machine learning for classification tasks. It is a vector that defines the decision boundary between different classes in a dataset. Support vectors are the data points that lie closest to the decision boundary and are used to define the optimal separating hyperplane. In summary, the main difference is that a position vector represents a point's location in space, while a support vector is used in machine learning for classification tasks.
Source:AI generated from FAQ.net
What is the rule for vector addition in vector geometry?
In vector geometry, the rule for vector addition is that the sum of two vectors is obtained by adding their corresponding componen...
In vector geometry, the rule for vector addition is that the sum of two vectors is obtained by adding their corresponding components. This means that if we have two vectors, A = (a1, a2) and B = (b1, b2), then the sum of these two vectors, A + B, is (a1 + b1, a2 + b2). This rule can be extended to vectors in higher dimensions as well, where the sum of two vectors is obtained by adding their corresponding components.
Source:AI generated from FAQ.net
Why is the direction vector a support vector and derivative?
The direction vector is a support vector because it provides the direction in which a function is changing. It is also a derivativ...
The direction vector is a support vector because it provides the direction in which a function is changing. It is also a derivative because it represents the rate of change of the function in that direction. In other words, the direction vector gives us the slope of the function in a specific direction, making it both a support vector and a derivative.
Is the zero vector perpendicular or parallel to every vector?
The zero vector is perpendicular to every vector. This is because the dot product of the zero vector with any other vector is zero...
The zero vector is perpendicular to every vector. This is because the dot product of the zero vector with any other vector is zero, which is the definition of perpendicularity in the context of the dot product. On the other hand, the zero vector is not parallel to any vector, as it has no direction and therefore cannot be parallel to any non-zero vector.
Source:AI generated from FAQ.net
Why can't vector b be a multiple of vector a?
Vector b cannot be a multiple of vector a because for two vectors to be multiples of each other, they must be parallel and have th...
Vector b cannot be a multiple of vector a because for two vectors to be multiples of each other, they must be parallel and have the same direction. If vector b were a multiple of vector a, it would mean that the two vectors are parallel and have the same direction. However, if vector b is a multiple of vector a, it would imply that vector b lies on the same line as vector a, which is not necessarily the case. Therefore, vector b cannot be a multiple of vector a.
Source:AI generated from FAQ.net
How can one determine the normal vector if only the position vector and a direction vector are given?
To determine the normal vector when only the position vector and a direction vector are given, one can first find a vector perpend...
To determine the normal vector when only the position vector and a direction vector are given, one can first find a vector perpendicular to the direction vector by taking the cross product of the direction vector with any other vector. This new vector will be perpendicular to both the direction vector and the normal vector. Then, one can find the normal vector by taking the cross product of the position vector with the vector found in the previous step. This will give the normal vector to the plane defined by the position vector and the direction vector.
What is the vector that has seven times the magnitude of vector 110, if the vector is 710?
The vector that has seven times the magnitude of vector 110 is 770. This is because when a vector is multiplied by a scalar, the m...
The vector that has seven times the magnitude of vector 110 is 770. This is because when a vector is multiplied by a scalar, the magnitude of the vector is multiplied by that scalar. Therefore, 110 multiplied by 7 equals 770
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Lines Ray And Angles Grade 4 Worksheets The protractor has two scales – an inner scale and an outer scale. The inner scale is used for angles that open to the right, and the outer scale is used for angles that open to the left. You can use printable exercises to help students practice reading protractor charts. These exercises can also be used to help students determine the dimensions of angles in one-degree increments.
Students can practice using a protractor on the worksheets As a result, you may want to consider hiring a tutor or enrolling your child in a maths tutoring course initial side of a line is the straight line, and the second side is the angle's other side.
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proportion
Proportion is a statement that two ratios are equivalent,
written as a:b = c:d, as in the statement
2:3 = 6:9. Two functions are proportional
if for all x, f (x) = kg(x),
where k is the constant of proportionality.
If the value of k is not known, the statement may be written f (x) – g(x).
In the ratio-equation or proportion a:b = c:d, a and d are referred to as the outer terms, and b and c as the inner terms of the proportion. If the outer terms are interchanged and the inner terms are unchanged, the proportion remains valid. Inner terms can only be interchanged with outer terms if this is done simultaneously on both sides. The sides of a proportion may be interchanged. The product of the inner terms is equal to the product of the outer terms. a:b = c:d implies a × d = b × c.
In the relation a:b = c:d, d is called the fourth proportional. If a, b, and c are given, d can be calculated: d – (b × c)/a.
Suppose in a proportion with equal inner terms, a:b = b:c, c is to be calculated for given a and b. c is called the third proportional with respect to a and b. We have cb 2/a.
In the proportion a:m = m:d, m is the mean proportional or geometric mean of a and d; m = √(ad).
If a:b = c:d, we have a/b = c/d. We can add ± to each side of this equation. Thus
a/b ± = c/d ±, so that (a ± b)/b = (c ± d)/d.
That is, if a:b = c:d, then a ± b:b = c ± d:d. These are known as derived proportions.
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The base of a triangle is divided into three equal parts. If \[{t_1},{t_2},{t_3}\] are the tangents of the angles subtended by these parts at the opposite vertex, prove that\[\left( {\dfrac{1}{{{t_1}}} + \dfrac{1}{{{t_2}}}} \right)\left( {\dfrac{1}{{{t_2}}} + \dfrac{1}{{{t_3}}}} \right) = 4\left( {1 + \dfrac{1}{{{t_2}^2}}} \right)\].
Hint: To solve this question, we first make two lines from vertex to the base such that they divide it into three equal parts. Then we apply \[m:n\] rule in any two of the triangles thus formed. In this way we get two equations. We will divide them. Then we simplify it further to reach the given equation. Formula used: If \[\alpha \] and \[\beta \] are any two angles, then cot function of their sum is given as \[cot(\alpha + \beta ) = \dfrac{{\cot \alpha \cot \beta - 1}}{{cot\beta + cot\alpha }}\]. According to m-n theorem if \[m\] and \[n\] are the length of base of two triangles and \[\alpha \] and \[\beta \] are the angles of vertex, then \[\left( {m{\text{ }} + {\text{ }}n} \right)cot\theta = mcot\alpha-ncot{\beta} \]
Note: We have to be very careful while drawing the diagram as our whole solution is based on it. We also have to be careful while doing the calculation as it is lengthy and there are great chances of us missing out any term. The above formulas must be learnt as they make such problems easy to solve.
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Median, Altitude, Perpendicular Bisector Given a Graph
Description: Your students will work through 12 questions calculating the median, altitude and perpendicular triangle of line segments in triangles.
This activity uses graphs and scaffolded steps to help with calculations.
Students will count the rise and run on the graph to fill in the slope.
RESOURCE INCLUDES
13 questions
1 multiple choice match terminology to image
4 median of a line segment
4 altitude of a line segment
4 perpendicular bisector
self-checking
no-prep
paperless
digital internet activity
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Definition
A circle is characterized as the locus of a point undergoing planar motion, whereby its distance from a fixed point, termed the center, remains invariant. The constant distance is precisely identified as the radius of the circle.
Basic Theorems and Results of Circles
Concentric circles: Circles having same centre.
Congruent circles: Iff their radii are equal.
Congruent arcs: Iff they have same degree measure at the centre.
Theorem 1
If two arcs of a circle (or of congruent circles) are congruent, the corresponding chords are equal. Converse : If two chords of a circle are equal then their corresponding arcs are congruent.
Equal chords of a circle (or of congruent circles) subtend equal angles at the centre. Converse : If the angle subtended by two chords of a circle (or of congruent circle) at the centre are equal, the chords are equal.
Theorem 2
The perpendicular from the centre of a circle to a chord bisects the chord. Converse : The line joining the mid point of a chord to the centre of a circle is perpendicular to the chord.
Perpendicular bisectors of two chords of a circle intersect at tis centre.
Theorem 3
There is one and only one circle passing through three non collinear points.
If two circles intersects in two points, then the line joining the centres is perpendicular bisector of common chords.
Theorem 4
Equal chords of a circle (or of congruent circles) are equidistant from the centre. Converse: Chords of a circle (or of congruent circles) which are equidistant from the centre are equal.
If two equal chords are drawn from a point on the circle, then the centre of circle will lie on angle bisector of these two chords.
Of any two chords of a circle larger will be near to centre.
Theorem 5
The degree measure of an arc or angle subtended by an arc at the center is double the angle subtended by it at any point of alternate segment.
Angle in the same segment of a circle are equal.
The angle in a semi circle is right angle. Converse: The arc of a circle subtending a right angle in alternate segment is semi circle.
Theorem 6
An angle subtended by a minor arc in the alternate segment is acute and any angle subtended by a major arc in the alternate segment is obtuse.
Theorem 7
If a line segment joining two points subtends equal angles at two other points lying on the same side of the line segment, the four points are concyclic, i.e. lie on the same circle.
Cyclic Quadrilaterals
A quadrilateral is called a cyclic quadrilateral if its all vertices lie on a circle.
Theorem 1
The sum of either pair of opposite angles of a cyclic quadrilateral is 180º.
OR
The opposite angles of a cyclic quadrilateral are supplementary.
Converse : If the sum of any pair of opposite angle of a quadrilateral is 180º, then the quadrilateral is cyclic.
Theorem 2
If a side of a cyclic quadrilateral is produced, then the exterior angle is equal to the interior opposite angle.
Theorem 3
The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic.
Theorem 4
If two sides of cyclic quadrilateral are parallel then the remaining two sides are equal and the diagonals are also equal.
OR
A cyclic trapezium is isosceles and its diagonals are equal.
Converse : If two non-parallel sides of a trapezium are equal then it is cyclic.
OR
An isosceles trapezium is always cyclic.
Theorem 5
The bisectors of the angles formed by producing the opposite sides of a cyclic quadrilateral (provided they are not parallel), intersect at right angle.
Tangents to a Circle
Theorem 1
A tangent to a circle is perpendicular to the radius through the point of contact.
Converse: A line drawn through the end point of a radius and perpendicular to it is a tangent to the circle.
Theorem 2
If two tangents are drawn to a circle from an external point, then:
They are equal.
The subtend equal angles at the centre.
They are equally inclined to the segment, joining the centre to that point.
Theorem 3
If two chords of a circle intersect inside or outside the circle when produced, the rectangle formed by the segments of one chord is equal in area to the rectangle formed by the two segments of the other chord
PA × PB = PC × PD
Theorem 4
If PAB is a secant to a circle intersecting the circle at A and B and PT is tangent segment, the PA × PB = PT2
OR
Area of the rectangle formed by the two segments of a chord is equal to the area of the square of side equal to the length of the tangent from the point on the circle.
Theorem 5
If a chord is drawn through the point of contact of tangent to a circle, then the angles which this chord makes with the given tangent are equal respectively to the angles formed in the corresponding alternate segments.
Converse
Standard equations of the circle
Central Form
If (h, k) is the centre and r is the radius of the circle then its equation is (x - h)2 + (y - k)2 = r2
Special Cases
(i) If centre is origin (0, 0) and radius is `r' then equation of circle is x2 + y2 = r2 and this is called the standard form.
The line will lie between these circle if centre of the circles lie on opposite sides of the line, i.e.
Line wouldn't touch or intersect the circles if,
or or
Hence common values of `a' are .
Example 8. The equation of a circle whose centre is (3, -1) and which cuts off a chord of length 6 on the line
2x - 5y + 18 = 0. Sol.
Let AB(= 6) be the chord intercepted by the line 2x - 5y + 18 = 0 from the circle and let CD be the perpendicular drawn from centre (3, - 1) to the chord AB.
i.e.,
Therefore, . Hence required equation is
Example 9. The area of the triangle formed by line joining the origin to the points of intersection(s) of the line and circle x2 + y2 = 10 is
Sol.
Length of perpendicular from origin to the line is
Radius of the given circle
.
Thus area of
➢ Equation of the tangent (T = 0)
Tangent at the point (x1, y1) on the circle is .
The tangent at the point (acos t, asin t) on the circle is
The point of intersection of the tangents at the points and is
The equation of tangent at the point (x1, y1) on the circle is
If line y = mx + c is a straight line touching the circle , then and contact points are or and equation of tangent is
The equation of tangent with slope m of the circle is
Note : To get the equation of tangent at the point (x1, y1) on any curve we replace xx1 in place of x2, yy1 in place of in place of in place of y, in place of xy and c in place of c.
➢ Length of tangent :
The length of tangent draw from point (x1, y1) out side the circle
is,
Note : When we use this formula the coefficient of x2 and y2 must be 1.
➢ Equation of Pair to tangents (SS1 = T2)
Let the equation of circle and is any point outside the circle.
From the point we can draw two real and distinct tangent PQ & PR and combine equation of pair of
tangents is or
Example 10.Let A be the centre of the circle x2 + y2 - 2x - 4y - 20 = 0 and B(1, 7) and D(4, -2) are points on the circle then, if tangents be drawn at B and D, which meet at C, then area of quadrilateral ABCD is
Sol.
or y = 7 ...(i)
Tangent at is ...(ii)
Solving (i) and (ii), C is (16, 7)
Area ABCD = AB × BC = 5 × 15 = 75 units.
Normal of Circle
Normal at a point of the circle is the straight line which is perpendicular to the tangent at the point of contact and passes through the centre of circle.
(a) Equation of normal at point (x1, y1) or circle is
(b) The equation of normal on any point (x1, y1) of circle is
(c) If is the equation of the circle then at any point `t' of this circle (a cos t, a sin t), the equation of normal is x sin t - y cos t = 0.
Let AB be the chord which is bisected by x-axis at a point M. Let its co-ordinates be M(h, 0).
and Equation of chord AB is T = S1
hx + 0 - (x + h) - (y + 0) = h2 + 0 - ah - 0
Since its passes through (a, b/2) we have
Now there are two chords bisected by the x-axis, so there must be two distinct real roots of h.
B2 - 4AC > 0
⇒ - 4.1. > 0
⇒ a2 > 2b2.
Director Circle
The locus of point of intersection of two perpendicular tangents to a circle is called director circle. Let P(h, k) is the point of intersection of two tangents drawn on the circle x2 + y2 = a2. Then the equation of the pair of tangents is SS1 = T2.
i.e. (x2 + y2 - a2) (h2 + k2 - a2) = (hx + ky - a2)2
As lines are perpendicular to each then, coefficient of x2 + coefficient of y2 = 0.
⇒ [(h2 + k2 - a2) - h2] + [(h2 + k2 - a2) - k2] = 0
⇒ h2 + k2 = 2a2
locus of (h, k) is x2 + y2 = 2a2 which is the equation of the director circle.
director circle is a concentric circle whose radius is times the radius of the circle.
Example 15. Let P be any moving point on the circle x2 + y2 - 2x = 1, from this point chord of contact is drawn w.r.t the circle x2 + y2 - 2x = 0. Find the locus of the circumcentre of the triangle CAB, C being centre of the circle.
Sol. The two circles are (x - 1)2 + y2 = 1 ........(i)
(x - 1)2 + y2 = 2 ........(ii)
So the second circle is the director circle of the first. So
Now circumcentre of the right angled triangle CAB would lie on the mid point of AB
So let the point be
Now, CM = CB sin 45º =
So, (h - 1)2 + k2 =
So, locus of M is (x - 1)2 + y2 = .
Pole and Polar
Let any straight line through the given point A(x1 , y1) intersect the given circle S = 0 in two points P' and Q and if the tangent of the circle at P and Q meet at the point R then locus of point R is called polar of the point A and point A is called the pole, with respect to the given circle.
Let PQR is a chord which passes through the point P(x1, y1) which intersects the circle at point Q and R and the tangents are drawn at points Q and R meet at point S(h, k) then equation of QR the chord of contact is
x1h + y1k = a2
locus of point S(h, k) is xx1 + yy1 = a2 which is the equation of the polar.
Notes :
(i) The equation of the polar is the T = 0, so the polar of point (x1, y1) w.r.t. circle
(ii) If point is outside the circle then equation of polar and chord of contact is same. So the chord of contact is polar.
(iii) If point is inside the circle then chord of contact does not exist but polar exists.
(iv) If point lies on the circle then polar, chord of contact and tangent on that point are same.
(v) If the polar of P w.r.t. a circle passes through the point Q, then the polar of point Q will pass through P and hence P and Q are conjugate points of each other w.r.t. the given circle.
(vi) If pole of a line w.r.t. a circle lies on second line. Then pole of second line lies on first line and hence both lines are conjugate lines of each other w.r.t. the given circle.
➢ Pole of a given line with respect to a circle.
To find the pole of a line we assume the coordinates of the pole then from these coordinates we find the polar. This polar and given line represent the same line. Then by comparing the coefficients of similar terms we can get the coordinates of the pole. The pole of lx + my + n = 0.
w.r.t. circle x2 + y2 = a2 will be .
Family of Circles
The equation of the family of circles passing through the points of intersection of two circles S1 = 0 and S2 = 0 is ; S1 + KS2 = 0 ().
The equation of the family of circles passing through the point of intersection of a circle S = 0 and line L = 0 is given by S + KL = 0.
The equation of a family of circles passing through two given points (x1 , y1) and (x2 , y2) can be written in the form (x- x1) (x - x2) + (y - y1) (y - y2) + K = 0 where K is a parameter.
Equation of circle circumscribing a quadrilateral whose side in order are represented by the line L1=0, L2 = 0, L3 = 0 and L4 = 0 are L1L3 + λL2L4 = 0 provided coefficient of x2 = coefficient of y2 and coefficient of xy = 0.
Since it touches the line x + 2y = 0, hence, Radius = Perpendicular distance from centre to the line.
i.e.,
cannot be possible in case of circle. So λ = 1. Thus, we get the equation of circle.
Direct and Transverse Common Tangents
Let two circles having centre C1 and C2 and radii, r1 and r2 and C1C2 is the distance between their centres then
(a) Both circles will touch
(i) Externally : If c1c2 = r1 + r2 i.e. the distance between their centres is equal to sum of their radii and point P divides C1C2 in the ratio r1 : r2 (internally). In this case there will be three common tangents.
(ii) Internally : If C1C2 = |r1 - r2| i.e. the distance between their centres is equal to difference between their radii and point P divides C1C2 in the ratio r1 : r2externally and in this case there will be only one common tangent.
(b) The circles will intersect : when |r1 - r2| < C1C2 < r1 + r2 in this case there will be two common tangents.
(c) The circles will not intersect :
(i) One circle will lie inside the other circle if C1C2 < |r1 - r2| In this case there will be no common tangent.
(ii) When circle are apart from each other the C1C2 > r1 + r2 and in this case there will be four common tangents. Lines PQ and RS are called transverse or indirect or internal common tangents and these lines meet line C1C2 on T1 and T1 divides the line C1C2 in the ratio r1 : r2 internally and lines AB & CD are called direct or external common tangents.
Let C1 and C2 be the centre of circle (i) and (ii), respectively and r1 and r2 be their radii, then
, ,
Here we find the two circles touch each other internally or externally.
For touch, |C1C2| = |r1 ± r2| or
On squaring
or
Again squaring, or or
The Angle of Intersection of Two Circles
Definition: The angle between the tangents of two circles at the point of intersection of the two circles is called angle of intersection of two circles. If two circles are
and θ is the angle between them
then or
Here a1 and a2 are the radii of the circles and d is the distance between their centres.
If the angle of intersection of the two circles is a right angle then such circles are called "Orthogonal circles" and conditions for the circles to be orthogonal is
Radical Axis of the Two Circles (S1 - S2 = 0)
➢ Definition : The locus of a point, which moves in such a way that the length of tangents drawn from it to the circles are equal and is called the radical axis.
It two circles are -
Let P(h, k) is a point and PA, PB are length of two tangents on the circles from point P. Then from definition : = or
locus of (h, k)
⇒ ⇒S1 - S2 = 0
which is the equation of radical axis.
Notes :
(i) To get the equation of the radical axis first of all make the coefficient of x2 and y2 = 1
(ii) If circles touch each other then Radical axis is the common tangent to both the circles.
(iii) When the two circles intersect on real points then common chord is the Radical axis of the two circles.
(iv) The Radical axis of the two circles is perpendicular to the line joining the centre of two circle of two circles but not always pass through mid point of it.
(v) The Radical axis of three circles (Taking two at a time) meet on a point.
(vi) If circles are concentric then the Radical axis does not always exist but if circles are not concentric then Radical axis always exists. (vii) If two circles are orthogonal to the third circle then Radical axis of both circles passes through the centre of the third circle.
(viii) A system of circle, every pair of which have the same radical axis, is called a coaxial system of circles.
➢ Radical centre
The radical centre of three circles is the point form which length of tangents on three circles are equal i.e. the point of intersection of radical axis of the circles is the radical centre of the circles. To get the radical axis of three circles S1 = 0, S2 = 0, S3 = 0 we have to solve any two S1 - S2 = 0, S2 - S3 = 0, S3 - S1 = 0
Notes :
(i) The circle with centre as radical centre and radius equal to the length of tangent from radical centre to any of
the circle, will cut the three circles orthogonally.
(ii) If three circles are drawn on three sides of a triangle taking them as diameter then its orthocentre will be its radical centre.
(iii) Locus of the centre of a variable circle orthogonal to two fixed circles is the radical axis between the two fixed circles.
(iv) If two circles are orthogonal, then the polar of a point `P' on first circle w.r.t. the second circle passes through the point Q which is the other end of the diameter through P. Hence locus of a point which moves such that its polars w.r.t. the circles, S1 = 0, S2 = 0 & S3 = 0 are concurrent is a circle which is orthogonal to all the three circles.
Example 18. A and B are two fixed points and P moves such that PA = nPB where Show that locus of P is a circle and for different values of n all the circles have a common radical axis.
Sol. Let and so PA = nPB
Hence locus of P is , which is a circle of different values of n.
Let n1 and n2 are two different values of n so their radical axis is x = 0 i.e. y-axis. Hence for different values of n the circle have a common radical axis.
Ans. The basic theorems and results of circles include the theorem on cyclic quadrilaterals, the concept of tangents to a circle, and the standard equations of a circle.
2. What is Theorem 4 related to circles?
Ans. Theorem 4 in the context of circles is not specified in the given article. It is recommended to refer to the article or textbook for the specific details of Theorem 4.
3. How are tangents related to circles?
Ans. Tangents to a circle are lines that intersect the circle at exactly one point. They are perpendicular to the radius drawn to the point of tangency. Tangents play a crucial role in understanding the properties and geometry of circles.
4. What are cyclic quadrilaterals?
Ans. Cyclic quadrilaterals are quadrilaterals whose four vertices lie on a single circle. The opposite angles of a cyclic quadrilateral are supplementary, meaning that they add up to 180 degrees.
5. What are the standard equations of a circle?
Ans. The standard equation of a circle in the coordinate plane is given by (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the coordinates of the center of the circle and r represents the radius. This equation helps in describing the position, size, and shape of a circle.
Document Description: Circle: Theorems and Equation for JEE 2024 is part of Mathematics (Maths) for JEE Main & Advanced preparation.
The notes and questions for Circle: Theorems and Equation have been prepared according to the JEE exam syllabus. Information about Circle: Theorems and Equation covers topics
like Definition, Basic Theorems and Results of Circles, Theorem 4, Cyclic Quadrilaterals, Tangents to a Circle, Standard equations of the circle and Circle: Theorems and Equation Example, for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Circle: Theorems and Equation.
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Elements of Geometry: Containing the First Six Books of Euclid, with a ...
equal to it and the angle at A is half the angle BGC, and the angle at D'half of the angle EHF; therefore the angle at A is equal to the angle at D.
PROP. XXVII. THEOR.
In equal circles, equal straight lines cut off equal arcs, the greater equal to the greater, and the less to the less.
Let ABC, DEF be equal circles, and BC, EF equal straight lines in them, which cut off the two greater arcs BAC, EDF, and the two less BGC, EHF: the greater BAC is equal to the greater EDF, and the less BGC to the less EHF.
Take K, L, the centres of the circles, and join BK, KC, EL, LF; and because the circles are equal, the straight lines from their centres are equal;
therefore BK, KC are equal to EL, LF; but the base BC is also equal to the base EF; therefore the angle BKC is equal (5. 1.) to the angle ELF: and equal angles stand upon equal (25. 3.) arcs, when they are at the centres; therefore the arc BGC is equal to the arc EHF. But the whole circle ABC is equal to the whole EDF; the remaining part, therefore, of the circumference, viz. BAC, is equal to the remaining part EDF.
PROP. XXVIII. THEOR.
In equal circles, equal arcs are subtended by equal straight lines. Let ABC, DEF be equal circles, and let the arcs BGC, EHF also be
equal; and join BC, EF: the straight line BC is equal to the straight line
EF.
Take K, L, the centres of the circles, and join BK, KC, EL, LF: and because the arc BGC is equal to the arc EHF, the angle BKC is equal (26. 3.) to the angle ELF: also because the circles ABC, DEF are equal, their radii are equal: therefore BK, KC are equal to EL, LF: and they contain equal angles; therefore the base BC is equal (1. 1.) to the base EF.
In a circle, the angle in a semicircle is a right angle; but the angle in a segment greater than a semicircle is less than a right ungle; and the angle in a segment less than a semicircle is greater than a right angle.
Let ABCD be a circle, of which the diameter is BC, and centre E; draw CA dividing the circle into the segments ABC, ADC, and join BA, AD, DC; the angle in the semicircle BAC is a right angle; and the angle in the segment ABC, which is greater than a semicircle, is less than a right angle; and the angle in the segment ADC, which is less than a semicircle, is greater than a right angle.
Join AE, and produce BA to F ; and because BE is equal to EA, the angle EAB is equal (3. 1.) to EBA: also, because AE is equal to EC, the angle EAC is equal to ECA; wherefore the whole angle BAC is equal to the two angles ABC, ACB. But FAC, the exterior angle of the triangle ABC, is also equal (25. 1.) to the two angles ABC, ACB; therefore the angle BAC is equal to the angle FAC, and each of them is therefore a right (7. Def. 1.) angle; wherefore the angle BAC in a semicircle is a right angle.
B
F
A
D
C
E
And because the two angles ABC, BAC of the triangle ABC, are together less (10. 1.) than two right angles, and BAC is a right angle, ABC must be less than a right angle; and therefore the angle in a segment ABC, greater than a semicircle, is less than a right angle.
Also because ABCD is a quadrilateral figure in a circle, any two of its opposite angles are equal (22. 3.) to two right angles; therefore the angles ABC, ADC are equal to two right angles; and ABC is less than a right angle; wherefore the other ADC is greater than a right angle.
COR. From this it is manifest, that if one angle of a triangle be equal to the other two, it is a right angle, because the angle adjacent to it is equal to the same two; and when the adjacent angles are equal, they are right angles.
PROP. XXX. THEOR.
IfLet the straight line EF touch the circle ABCD in B, and from the point B let the straight line BD be drawn cutting the circle: The angles which BD makes with the touching line EF shall be equal to the angles in the alternate segments of the circle: that is, the angle FBD is equal to the angle which is in the segment DAB, and the angle DBE to the angle in the segment BCD.
D
From the point B draw (Prob. 6. 1.) BA at right angles to EF, and take any point C in the arc BD, and join AD, DC, CB; and because the straight line EF touches the circle ABCD in the point B, and BA is drawn at right angles to the touching line, from the point of contact B. the centre of the circle is (19. 3.) in BA; therefore the angle ADB in a semicircle, is a right (29. 3.) angle, and consequently the other two angles, BAD, ABD, are equal (25. 1.) to a right angle; but ABF is likewise a right angle; therefore the angle ABF is equal to the angles BAD, ABD take from these equals the common angle ABD, and there will remain the angle DBF equal to the angle BAD, which is in the alternate segment of the circle. And because ABCD is a quadrilateral figure in a circle, the opposite angles BAD, BCD are equal (22. 3.) to two right angles; therefore the angles DBF, DBE, being likewise equal (6. 1.) to two right angles, are equal to the angles BAD, BCD; and DBF has been proved equal to BAD; therefore the remaining angle DBE is equal to the angle BCD in the alternate segment of the circle.
E
PROP. XXXI. THEOR.
B
F
If two straight lines within a circle cut one another, the rectangle contained by the segments of one of them is equal to the rectangle contained by the segments of the other.
Let the two straight lines, AC, BD, within the circle ABCD, cut one
another in the point E: the rectangle contained by AE, EC is equal to the rectangle contained by BE, ED.
If AC, BD pass each of them through the centre, so that È is the centre, it is evident that AE, EC, BE, ED, being all equal, the rectangle AE.EC is likewise equal to the rectangle BE.ED.
But let one of them BD pass through the
B
A
E
D
C
Ꭰ
centre, and cut the other AC, which does not pass through the centre, at right angles in the point E; then, if BD be bisected in F, F is the centre of the circle ABCD; join AF: and because BD, which passes through the centre, cuts the straight line AC, which does not pass through the centre at right angles, in E, AE, EC are equal (3. 3.) to one another; and because the straight line BD is cut into two equal parts in the point F, and into two unequal in the point E, BE.ED (5. 2.) +EF2=FB2=AF2. But AF2 = AE2
Next, Let BD, which passes through the centre, cut the other AC, which does not pass through the centre, in E, but not at right angles; then, as before, if BD be bisected in F, F is the centre of the circle. Join AF, and from F draw (Prob. 7. 1.) FG perpendicular to AC; therefore AG is equal (3. 3.) to GC; wherefore AE.EC+(5. 2.) ÈG2=AG2, and adding GF2 to both, AE.EC + EG2+GF2 = AG2+GF2. Now EG2+ GF2 = EF2, and AG2+GF2AF2; therefore AE.EC+EF2 = AF2-FB2 But FB2= BE.ED+(5. 2.) EF2, therefore AE.EC+EF2=BE.ED+EF2, and tak
ing EF2 from both, AE.EC=BE.ED.
=
Lastly, Let neither of the straight lines AC, BD pass through the centre: take the centre F, and through E, the intersection of the straight lines AC, DB, draw the diameter GEFH: and because, as has been shown, AE.EC-GE.EH, and BE.EDGE.EH; therefore AE.EC=BE.ED.
F
E
A
G
C
B
PROP. XXXII. THEOR.: and because the straight line AC is bisected in E, and produced to the point D, AD.DC+EC2 = ED2 (6. 2.). But EC= EB, therefore AD.DC+ EB2=ED2. Now ED2=(37.1.) EB2+BD2, because EBD is a right angle; therefore AD. DC+EB2 EB2+BD2, and taking EB2 from each, AD.DC=BD2.
But, if DCA does not pass through the centre of the circle ABC, take the centre E, and draw EF perpendicular (Prob. 7. 1.) to AC, and join EB, EC,ED: and because and produced to D (6. 2.), AD.DC + FC2 FD2; add FE2 to both, then AD.DC + FC2 +FE2 = FD2 + FE2. But (37. 1.) EC2 = FC2FE2, and ED2 = FD2+FE2, because DFE is a right angle; therefore AD.DC+ EC2 ED2. Now, because EBD is a right angle, ED2 EB2+BD2 EC2+BD2, and therefore, AD.DC + EC2 EC2+BD2, and AD.DC=BD2.
=
=
COR. 1. If from any point without a circle, there be drawn two straight lines cutting it, as AB, AC, the rectangles contained by the whole lines and the parts of them without the circle, are equal to one another, viz. BA.AE =CA.AF; for each of these rectangles is equal to the square of the straight line AD, which touches the circle.
COR. 2. It follows, moreover, that two tangents drawn from the same point are equal.
COR. 3. And since a radius drawn to the point of contact is perpendicular to the tangent, it follows that the angle included by two tangents,
drawn from the same point, is bisected by a line
drawn from the centre of the circle to that point; for this line forms the hypo
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Geometry I: Snow Leopards Look in Circles & Prisms
$42.00Price
A snow leopard's tail is thought to help it with balance and to wrap around itself for warmth in the cold habitat the snow leopard lives in. They love spending time in snowy areas to think about constructing and measuring polygons, measuring circles, solving for volume and surface area, and working with 2D and 3D shapes.
Students construct different types of triangles, quadrilaterals, regular polygons, and scaled polygons using specified restraints, and use a ruler and protractor to measure their sides and angles.
Students build and solve equations in order to solve for the area and circumference of circles.
Students build and solve equations in order to solve for the volume and surface area of prisms and pyramids.
Students identify the two-dimensional figure that is the result of slicing a three-dimensional solid with a plane.
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An Introduction to Geometry and the Science of Form: Prepared from the Most Approved Prussian Text-books
From inside the book
Results 1-5 of 36
Page xviii ... ABC DEF means that the triangle ABC is similar to DEF . AB means a square each side of which is equal to the line AB . To avoid the frequent repetition of the term right angle , the abbreviation R. A. is substituted . Thus 2 R. A. is ...
Page 60 ... TRIANGLES . 70. To construct a triangle which shall be equal to another triangle ABC , ( fig . 62. ) . 1. Let the 3 sides of ABC be given . Draw a straight line EF equal to one of the given sides , for example , BC . From E as a centre ...
Page 61 ... ABC , to be known . Draw a straight line EF = side BC . At point E make an angle DEF = CBA . Take ED BA , and draw DF . DEF is the triangle required . = That the triangles thus constructed are in each case equal to the given triangle ...
Page 65 ... triangles , ABC and ADC , but AEFC is composed of 4 triangles , each equal to ABC . 82. Remark . In a right triangle the side opposite to the right angle is called the hypothenuse . Now you will observe that the square constructed upon ...
Page 68 ... triangle , we may construct a regular 6 , 12 , 24 , & c . sided polygon in a circle , and by drawing tangents at the ... ABC , ( fig . 63. 2. ) Sol . Bisect the angles A and B by straight lines , which will intersect each other at O. From
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What is the measure of the reference angle for a angle?
What is Meant by the Reference Angle? In mathematics, the reference angle is defined as the acute angle and it is measuring less than 90 degrees. It is always the smallest angle, and it makes the terminal side of an angle with the x-axis.
How do you solve reference angles?
Choose a proper formula for calculating the reference angle:
0° to 90°: reference angle = angle ,
90° to 180°: reference angle = 180° – angle ,
180° to 270°: reference angle = angle – 180° ,
270° to 360°: reference angle = 360° – angle .
What is the measure of the reference angle for a degree angle?
So, if our given angle is 110°, then its reference angle is 180° – 110° = 70°. When the terminal side is in the third quadrant (angles from 180° to 270°), our reference angle is our given angle minus 180°. So, if our given angle is 214°, then its reference angle is 214° – 180° = 34°.
How do you find the reference angle step by step?
How to find the reference angle for degrees
Choose your initial angle – for example, 610°.
If your angle is larger than 360° (a full angle), subtract 360°. Keep doing it until you get an angle smaller than a full angle. This is the same as finding the modulo.
Which is the measure of the reference angle for 227 degrees?
What is the significance of the unit circle?
A unit circle is a circle with a radius of 1, and it is used to show certain common angles . Unit circle: Commonly encountered angles measured in radians and degrees.
What is the sine of a circle?
Verified by Expert. 1) Amplitud: in the unit circle, sine is defined as the length of the vertical projection of the (vetical axis) of the point that defines the angle. Whose maximum value is 1, which is the amplitud of the sine
| 677.169 | 1 |
Convert Points from Polar to Rectangular
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Hey, everyone. Now that we're familiar with both polar and rectangular coordinates, we're going to have to convert between the two and take a point given in polar coordinates as our theta converting it into its rectangular xy equivalent. Now this all just comes back down to triangles, something that you are already an expert on. So here, I'm going to break down for you exactly how to take a point given in polar coordinates and convert it into rectangular coordinates just using this triangle here. So let's go ahead and get started. Now looking at the point that we have on our graph here given in polar cod its five pi over three, we can go ahead and form a triangle with this giving our triangle a hypotenuse of five and an inner angle theta of pi over three. Then our side links here are X and Y the same way that we've seen before on the unit circle. But here, our hypotenuse is no longer just one, the same way that it was there. Now, we can dive a bit deeper here because if I set up a cosine equation here, I would see that the cosine of data of course, is equal to the adjacent side over the hypo nose. And specifically this triangle that would tell me that the cosine of that inner angle pi over three is equal to my side length X over my hypotenuse of five. Then I can do the same thing for a sine function, setting up my sine function sine of theta is equal to the opposite side, which in this case is Y over that hypotenuse value of five. Now I can go ahead and solve each of these for X and Y. So looking at my cosine equation here, if I go ahead and multiply both sides by five to cancel that out on the right side, I end up getting that five times the cosine of pi over three is equal to X. Then doing the same thing for my sine equation. Again, multiplying both sides by five here having that cancel on the right side. I see that five times the sine of pi over three is equal to Y. Now, if I actually multiply these out, I see that I get an X value of 5/2 and A Y value of five root 3/2. So looking at the point that I started with five pi over three in polar coordinates, I now have my point in rectangular coordinates as 5/2 5 root 3/2. Now this will work for any point given in polar coordinates when converting to rectangular, my X value is always going to be equal to R times the cosine of data and Y will always be equal to R times the sine of data. Now, this probably looks really familiar to you because from the unit circle, we saw that X was equal to the cosine of data and Y is equal to the sine of data. Now, we just have this R value to account for, but we're still doing this same exact thing just that R is no longer always equal to one. But now that we know how to find these rectangular points given a polar coordinate. Let's go ahead and get some practice here and work through some examples together. Here, we're given a point in polar coordinates negative three pi over six. Now, here we want to go ahead and plot this point on our polar coordinate system and then convert it into rectangular coordinates. So let's go ahead and start by plotting this on our polar coordinate system negative three pi over six. I locate my angle theta first pi over six along this line. But since I have a negative R value three, I'm going to count in the opposite direction three units and I end up with my point right here. Now that I have that point plotted, let's go ahead and convert this into rectangular coordinates. Now I know that my X value is going to be equal to R times the cosine of theta So X equals R cosine of theta here plugging in my values for R and theta, I get negative three times the cosine of pi over six. Now the cosine of pi over six is the square root of 3/2. So this is negative three times the square root of 3/2. Now I have my X value, let's find our Y value Y is going to, of course, be equal to R times the sine of the. So when we plug these values in here, that ends up giving me a negative three times the sine of that angle pi over six. Now the sign of pi over six is one half. So this is equal to negative three times one half which I can rewrite as being negative 3/2. So now I have my X and Y value negative three root 3/2 and negative 3/2. Now this is my point in rectangular coordinates. Now, in this problem, we're also asked to go ahead and plot this on the xy plane, which is actually going to be easier to do if we have these in their decimal form. So this negative three root 3/2 is going to be about negative 2.6 and negative 3/2 is of course just negative 1.5. So I can go ahead and use those decimals to plot that on my rectangular coordinate system. Now actually locating the X value negative 2.6. And then going down to my Y value of negative 1.5 I end up right here in quadrant three of my rectangular coordinate system. Now, here we see that these points are located in the exact same spot, which makes sense because these are just the exact same point but represented in different ways. One in polar and one in rectangular. Now let's look at one final example here, here we have the point given in polar coordinates, zero negative pi over six. Now again, here we want first plot this on our polar coordinate system. So locating my angle negative pi over six, which will end up being right along this line. Since I know that my R value is zero, I actually just stay right here at the pole. What we think of as being the origin in rectangular coordinates. So here is my point B now that we have that plotted on that polar coordinate system, let's go ahead and calculate our X and Y values and get this point in rectangular coordinates. Now I know that X is going to be equal to R times the cosine of theta which here my R value is zero. So this is zero times the cosine of negative pi over six but zero times anything is just zero. So this ends up giving me an X value of zero. Then for Y, if I also do the same thing here, R times the sine of the plugging in my values of R and theta I get zero times the sign of negative pi over six, which of course, again, zero times anything is still just zero. So I again get a Y value of zero then that gives me my point in rectangular cord zero comma zero, which we know is located at the origin, which I can go ahead and plot right here. So anytime we have an R value zero, that's always going to end up being the 0.00 in rectangular coordinates. And we again see that this is located in the exact same spot as it was on our polar coordinate system. Again, these are the same exact point represented in different ways. Now that we know how to convert points from polar to rectangular. Let's continue practicing. Thanks for watching and I'll see you in the next one.
2
Problem
Problem
Convert the point to rectangular coordinates.
(−2,−π4)(-2,-\frac{\pi}{4})(−2,−4π)
A
(2,−2)(\sqrt{2},-\sqrt2)(2,−2)
B
(−1,1) (-1,1)(−1,1)
C
(−2,2)(-\sqrt{2},\sqrt{2})(−2,2)
D
(−22,22)(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})(−22,22)
3
Problem
Problem
Convert the point to rectangular coordinates.
(4,π6)(4,\frac{\pi}{6})(4,6π)
A
(23,2)(2\sqrt{3},2)(23,2)
B
(43,4)(4\sqrt{3},4)(43,4)
C
(2,23)(2,2\sqrt{3})(2,23)
D
(2,3)(2,\sqrt{3})(2,3)
4
Problem
Problem
Convert the point to rectangular coordinates.
(−3,0)(-3,0)(−3,0)
A
(−3,0) (-3,0)(−3,0)
B
(0,−3) (0,-3)(0,−3)
C
(0,0) (0,0)(0,0)
D
(3,0) (3,0)(3,0)
5
Problem
Problem
Convert the point to rectangular coordinates.
(0,7π4)(0,\frac{7\pi}{4})(0,47π)
A
(0,74)(0,\frac74)(0,47)
B
(−4,0) (-4,0)(−4,0)
C
(0,0) (0,0)(0,0)
D
(0,7) (0,7)(0,7)
6
concept
Convert Points from Rectangular to Polar
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Hey, everyone, we just learned how to take points given in polar coordinates and convert them to rectangular coordinates. But we also need to be able to do the opposite take points given in rectangular coordinates and convert them to polar coordinates. Now, looking at this point given in rectangular coordinates here we see this sort of suspiciously familiar looking triangle that we've solved the time and time again. And that's all this is, it all just comes back down to triangles. So here we're going to keep using everything that we already know specific working with the Pythagorean theorem and the tangent function in order to convert points from their rectangular coordinates into polar coordinates. Now, here I'm going to walk you through how to do that step by step. So let's go ahead and get started. Now, looking at the point that we have here in rectangular coordinates, three comma four, we're able to form this triangle and from this rectangular coordinate 0.3 comma four, I know that this side LX is equal to three and my side length Y is equal to four. Now, since I already know X and Y it's just left a find R and theta, which will give me my point in polar coordinates. Now, let's start by finding R. Now looking at my R value here, this is the hypo news of my triangle, but I have my other two side links already. So I can find that hypo news by simply using the Pythagorean theorem, which tells me that C squared is equal to A squared plus B squared or specifically for this triangle R squared is equal to X squared plus Y squared. Now actually plugging in our values here, I get that R squared is equal to three squared plus four squared. Now adding those together gives me 25. So I have R squared is equal to 25 or completely solving that for R simply R is equal to five. Now that I have that R value, let's turn to finding the. Now looking at theta here, I know that I have my opposite side and my adjacent side, which means that I can set up a tangent function because I know that the tangent of the is equal to the opposite over the adjacent side or specifically for this angle for a point given in rectangular coordinates, the tangent of my angle theta is equal to Y over X. Now, again, just plugging in the values that I have here, I end up getting that the tangent of theta is equal to that Y value of four over that X value of three. Now in order to actually solve for theta. Here, I need to take the inverse tangent. So here theta will be equal to the inverse tangent of 4/3. Now, if we actually plug that inverse tangent into our calculator, we end up getting that theta is equal to about 53 degrees. And I now have both my R and my theta value. Now, I originally started with my point in rectangular coordinates, three comma four. And now I have my point in polar coordinates, five comma 53 degrees. Now these are the general equations that we're going to use to convert points from rectangular to polar coordinates. But we have to be really careful here because we end up taking an inverse tangent and the inverse tangent. Remember that this function is only defined over the intervals contained in quadrant one and quadrant four. And even though our point here was located in quadrant one, it won't always be. So I actually have a step by step process for you that will work no matter where your point is located. So let's go ahead and work through these examples together. Now first, we're given at this point here negative 40. And I want to go ahead and plot this point and then convert it into coordinates by following these steps here. Now starting with step one, we just want to go ahead and plot this point on a rectangular coordinate system. Now locating this point negative 40, I end up right here on my X axis for this 1st 0.8. Now with step one done, we move on to step two and calculating R. Now this is the same equation that we saw above but just already solved for R having taken the square root on both sides. So here plugging in my values, I get that R is equal to the square root of negative four squared plus zero squared. Now this ends up being the square root of 16, which is simply equal to four. So I have an R value here of four. Now, with our calculated, we move on to finding theta. Now, in order to find theta, we need to look at the location of our point and looking at where this point A is on my X axis because it's on an axis. I know that theta is going to be one of my quadrant angles and thinking specifically about where this point is located. Thinking about my angles here going from zero to pi radiance or 180 degrees. I know that my angle here is simply pi. So that gives me my value of theta pi and I now have my point in polar coordinates for comma pi. Now let's move on to our next example. Here we are specifically given the point negative one route three. Now we're going to restart our steps here starting from step number one, we're going to start out by plotting our point. Now plotting my point here, it's going to be more useful to know so that the square root of three is about 1.73 as a decimal. So plotting this point at negative 11.73 ends me up right here in quadrant two for point B. Now, at that point plotted, we want to go ahead and find R as we have before taking the square root of X squared plus Y squared. Now plugging these values in here, I end up getting the square root of negative one squared plus the square root of three squared. Now actually adding those together gives me the square root of four or simply two. So I have my R value of two. Now all that's left is to find theta. Again, we're looking at the location of our point and I see that my point B is located in quadrant two. So it's not on an axis. It's not in quadrant one or four as my original example was, but it is in quadrant two. So what do we do now? Well, we're gonna start off the same way that we did for our original angle that was in quadrant one by taking the inverse tangent of Y over X. So let's start there beta is equal to the inverse tangent of my Y value root three over my X value negative one. Now this simplifies to the inverse tangent of the negative square root of three, which from my knowledge of the unit circle or by simply typing this into a calculator, I will end up getting a negative pi over three for that inverse tangent of negative root three. But we're not done yet because let's think about where this point is located or where this angle is located. The negative square root of three is here in quadrant four. And that is not where my point B is located. So that wouldn't make any sense. Now, in order to make sure that this is located in the right place because our point is located here in quadrant two, we need to go ahead and add pi to that angle that we got from the inverse tangent. So this negative pi over three, I need to add pi to it. This ends up giving me two pi over three which thinking about where that angle is located that is located in quadrant two as it should be. So here I have my final value of two pi over three and I have this point. Now in polar coordinates, 22 pi over three. Now when converting points from rectangular to polar coordinates, we're going to start out the same way we're going to plot our point on our graph. And we're going to find R by using this equation here. Then when finding the, we need to pay attention to the location of our point and then we're good to go now that we know how to convert points from a rectangular to polar. Let's continue practicing. Thanks for watching and I'll see you in the next one.
7
example
Convert Points from Rectangular to Polar Example 1
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Here, we're asked to convert the point given in rectangular coordinates into polar coordinates. And the point that we're given here is three negative three. So let's go ahead and get started with our steps so that we can get this to polar coordinates. Now, step one tells us to go ahead and plot our point. So here on my rectangular coordinate system, I'm going to plot three negative three and I end up right down here in quadrant four. Now moving on to step two and actually finding our R value by taking the square root of X squared plus Y squared. Now plugging in my values here three and negative three. This gives me that R is equal to the square root of three squared plus negative three squared. Now three squared and negative three squared are both nine. So this gives me the square root of nine plus nine or the square root of 18. Now we can simplify this further using our radical rules in order to get that R is equal to three times the square root of two. Now let's go ahead and move on to step number three in finding the. Now here we want to pay attention to where our point is located and it is located in quadrant four. So that tells me that I want to find theta by taking the inverse tangent of Y over X. Now actually setting that up and plugging my values in, I get that theta will be equal to the inverse tangent of my Y value negative three over my X value three. Now this simplifies to give me the inverse tangent of negative one. And if I plug this into my calculator, this will give me negative pi over four for my value of data. So here I have my in polar coordinates with that R value three root two comma that beta value negative pi over four. So this is my final answer here three root two negative pi over four. Now something that you might be thinking about here if you chose to calculate your inverse tangent of Y over X differently and maybe not using a calculator. But rather the unit circle you may have thought isn't this angle also seven pi over four. And that's true. If I represented this point as three root to seven pi over four, that would still be correct. The reason that we got negative pi over four is because of the restrictions placed on our inverse tangent function. But remember that there are always multiple ways to represent the same point in polar coordinates. So if you were to have three root 27 pi over four, that would still be correct. And even if you tried to take that point back to rectangular coordinates, you would still end up with this 0.3 negative three. So let me know if you have any questions. Thanks for watching and I'll see you in the next one.
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How Many Sides Does a Polygon Have? Unveil the Mystery!
A polygon has multiple sides. The number of sides varies depending on the shape.
Have you ever wondered how many sides a polygon has? The answer is simple – a polygon is a closed shape with three or more sides. From triangles with three sides to polygons with countless sides, these shapes are found everywhere in nature and man-made structures.
Understanding the characteristics of polygons can help in various fields, from geometry to architecture. In this blog post, we will explore the fascinating world of polygons, their properties, and real-world applications. Let's dive in and unravel the mystery of these versatile geometric figures.
The Basic Definition Of A Polygon
A polygon is a two-dimensional shape that is formed by straight lines and is closed. It is a geometric figure that consists of a finite number of line segments connected end to end to form a closed shape.
Characteristics Of Polygons
Polygons have several defining characteristics:
Straight sides: All sides of a polygon are straight lines.
Closed shape: A polygon is a closed figure, which means that it has no openings.
Fixed number of sides: A polygon has a specific number of sides, with a minimum of three.
Interior angles: The interior angles of a polygon add up to a specific value, which varies based on the number of sides.
No intersecting sides: The sides of a polygon do not intersect with each other.
Polygons Versus Non-polygons
The distinction between polygons and non-polygons is clear:
Polygons
Non-Polygons
Have straight sides
May have curved sides
Are closed shapes
Can be open shapes
Have a fixed number of sides
Can have an indefinite number of sides
Classifying Polygons By Sides
Discovering the number of sides in a polygon involves classifying them based on their unique characteristics and shapes. Polygons can have varying numbers of sides, with common examples being triangles, quadrilaterals, pentagons, and hexagons. Understanding these distinctions helps in identifying and working with different types of polygons effectively.
Classifying Polygons by Sides When we talk about polygons, we are referring to closed figures with three or more straight sides. Polygons can have varying numbers of sides, and it is essential to classify them based on the number of sides they have. We can classify polygons into different categories, including triangles, quadrilaterals, and polygons with more than four sides. Below, we will discuss these categories in more detail.
Triangles: The Simplest Polygons
Triangles are the simplest polygons as they have only three sides. There are different types of triangles, including equilateral, isosceles, and scalene triangles. An equilateral triangle has three equal sides, while an isosceles triangle has two equal sides. A scalene triangle has no equal sides. Triangles are fundamental shapes, and they play a crucial role in the field of geometry.
Quadrilaterals And Beyond
Quadrilaterals are polygons with four sides, and they come in different shapes, including squares, rectangles, rhombuses, and trapezoids. A square has four equal sides and four right angles, while a rectangle has four right angles but not necessarily equal sides. A rhombus has four equal sides but no right angles, while a trapezoid has one pair of parallel sides. Polygons with more than four sides include pentagons, hexagons, heptagons, octagons, and nonagons. A pentagon has five sides, a hexagon has six sides, a heptagon has seven sides, an octagon has eight sides, and a nonagon has nine sides. These polygons also have unique properties and characteristics, making them essential in geometry.
Naming Polygons Up To 12 Sides
We can also name polygons based on the number of sides they have. For instance, a polygon with ten sides is known as a decagon, while one with twelve sides is called a dodecagon. Naming polygons makes it easier to classify them and study their properties. Below is a table showing the names of polygons up to twelve sides.
Number of Sides
Name of Polygon
3
Triangle
4
Quadrilateral
5
Pentagon
6
Hexagon
7
Heptagon
8
Octagon
9
Nonagon
10
Decagon
11
Hendecagon
12
Dodecagon
In conclusion, classifying polygons based on the number of sides they have is essential in the field of geometry. It enables us to understand their properties and characteristics, making it easier to study them. From triangles to polygons with twelve sides, each category has unique properties that make them stand out.
The Infinite Possibility Of Sides
Explore the endless world of polygons and their sides. Uncover the answer to the intriguing question: how many sides does a polygon have? Delve into the fascinating realm of geometry and discover the infinite possibilities that lie within these multifaceted shapes.
Beyond Dodecagons
A polygon is a geometric shape that consists of straight lines connected to form a closed figure. While most of us are familiar with common polygons like triangles, rectangles, and pentagons, the possibilities for the number of sides in a polygon are practically infinite. Beyond the well-known shapes, there exists a vast world of polygons with varying numbers of sides, each with its unique properties and characteristics.
One notable example is the dodecagon, a polygon with twelve sides. Its symmetrical structure and equal angles make it a popular choice in design and architecture. However, the exploration of polygons doesn't stop at twelve sides. In fact, mathematicians have discovered polygons with hundreds, thousands, and even millions of sides.
The concept of polygons with a large number of sides might seem abstract or even incomprehensible at first. After all, how can we visualize a shape with thousands of sides? To better understand this, let's consider the idea of a regular polygon, where all sides and angles are equal.
Imagine a regular polygon with a high number of sides, such as a thousand. As the number of sides increases, the polygon starts to resemble a circle more closely. In fact, as the number of sides approaches infinity, the polygon becomes indistinguishable from a circle. This fascinating connection between polygons and circles showcases the infinite potential of polygons.
When Does A Polygon Stop Being A Polygon?
While the possibilities for the number of sides in a polygon are limitless, there is a point where a shape stops being considered a polygon. To be classified as a polygon, a shape must meet certain criteria:
The shape must be two-dimensional, lying on a flat plane.
It must consist of straight lines.
The lines must be connected to form a closed figure.
Each line must intersect exactly two other lines, with no intersecting lines within the shape.
When any of these criteria are not met, the shape is no longer considered a polygon. For example, a curved line or a shape with intersecting lines would not be classified as a polygon. The definition of a polygon provides a clear boundary for what can be considered a polygon and what cannot.
Understanding the infinite possibility of sides in a polygon and the criteria for classification allows us to explore the diverse world of geometric shapes. By expanding our knowledge of polygons, we gain a deeper appreciation for the intricacies of mathematics and the beauty found in the seemingly simplest of shapes.
Regular Vs. Irregular Polygons
When it comes to polygons, they can be classified into two main categories: regular and irregular. Regular polygons have equal sides and equal angles, while irregular polygons have sides and angles of varying lengths and measures. Let's explore the characteristics of these two types of polygons in more detail.
Equilateral And Equiangular Regularity
Regular polygons are known for their symmetry and uniformity. One type of regular polygon is the equilateral polygon, where all sides are of equal length. An equilateral triangle is a prime example, with all three sides measuring the same. Other examples include the square, with four equal sides, and the regular pentagon, with five equal sides.
Equiangular regular polygons, on the other hand, have equal angles. Each interior angle in these polygons is the same. The equilateral triangle is also equiangular, with each angle measuring 60 degrees. The square is another equiangular regular polygon, with all angles measuring 90 degrees. Regular hexagons, octagons, and other polygons can also fall into this category.
The Diversity Of Irregular Shapes
Irregular polygons, as the name suggests, lack the uniformity found in regular polygons. These shapes can have sides of different lengths and angles of varying measures. The irregularity of these polygons leads to a vast array of shapes and sizes.
Some examples of irregular polygons include trapezoids, which have one pair of parallel sides, and parallelograms, which have opposite sides that are parallel and equal in length. Quadrilaterals such as rectangles and rhombuses can also be considered irregular polygons due to their varying angles.
Irregular polygons offer endless possibilities in terms of shape and design. Their unique characteristics allow for more creativity and flexibility in various applications, from architecture to art.
Examples of Regular and Irregular Polygons
Regular Polygons
Irregular Polygons
Equilateral Triangle
Trapezoid
Square
Parallelogram
Regular Pentagon
Rectangle
Regular Hexagon
Rhombus
Understanding the distinction between regular and irregular polygons is essential in geometry and various real-world applications. Regular polygons possess a sense of order and symmetry, while irregular polygons offer versatility and unique design possibilities.
The Role Of Angles In Polygons
Polygons are fascinating geometric shapes that are all around us. Understanding the angles within polygons is crucial to understanding their properties and characteristics. Let's delve into the role of angles in polygons and explore the Interior Angle Sum Property and the secrets of Exterior Angles.
Interior Angle Sum Property
In a polygon, the sum of its interior angles can be determined using a simple formula. For a polygon with n sides, the sum of its interior angles is equal to (n-2) 180 degrees. This property is essential in calculating individual interior angles and understanding the overall structure of polygons.
Exterior Angles And Their Secrets
The exterior angles of a polygon play a crucial role in understanding its overall shape and characteristics. Each exterior angle is supplementary to the interior angle adjacent to it. This means that the sum of each exterior angle and its corresponding interior angle is always 180 degrees. Understanding the secrets of exterior angles provides valuable insights into the symmetry and properties of polygons.
Mathematical Formulas Involving Polygons
Calculating The Sum Of Interior Angles
A polygon is a closed shape with straight sides. The sum of interior angles in a polygon can be calculated using the formula: Sum = (n – 2) 180° where n represents the number of sides.
Finding The Number Of Diagonals
The number of diagonals in a polygon can be found using the formula: Diagonals = n (n – 3) / 2 where n is the number of sides.
Polygons In The Real World
Exploring the fascinating world of polygons reveals their omnipresence in various aspects of our lives. From architectural marvels to nature's geometric preferences, polygons play a vital role in shaping our surroundings and natural environment.
Architectural Marvels
In architecture, polygons are fundamental in creating iconic structures. Skyscrapers, bridges, and domes often incorporate polygons to achieve stability and aesthetic appeal.
Nature's Geometric Preferences
Nature showcases polygons in intricate patterns of leaves, snowflakes, and beehives. The symmetry and efficiency of these natural formations highlight the beauty and functionality of polygons in the real world.
Credit: socratic.org
Exploring Polygons Through Technology
Embark on a journey to explore the world of polygons using innovative technology. Discover the answer to the intriguing question: "How many sides does a polygon have? " Engage in interactive learning experiences to gain a deeper understanding of polygonal shapes.
Computer Graphics And Polygons
Polygons play a crucial role in computer graphics.
The Role Of Polygons In 3d Modeling
Polygons serve as building blocks in 3D modeling. Polygons are used to create realistic 3D objects. In 3D models, polygons define the shape and surface. They are essential for rendering intricate structures. Polygons are fundamental elements in virtual environments. Polygons are the foundation of digital imagery. Computer graphics heavily rely on polygon manipulation. By understanding polygons, we enhance digital creativity. Polygons drive advancements in animation and design. Technology continues to push the boundaries of polygon usage.
Credit:
Frequently Asked Questions
How Many Sides Does A Polygon Have?
A polygon can have any number of sides, as long as it has at least three sides. The number of sides in a polygon determines its shape and name. For example, a triangle has three sides, a quadrilateral has four sides, and a pentagon has five sides.
The more sides a polygon has, the more complex its shape becomes.
What Is The Minimum Number Of Sides A Polygon Can Have?
The minimum number of sides a polygon can have is three. A polygon with three sides is called a triangle. Triangles are the simplest form of polygons and are made up of three line segments connected end to end. Triangles come in different types, such as equilateral, isosceles, and scalene, depending on the lengths of their sides.
Can A Polygon Have An Infinite Number Of Sides?
No, a polygon cannot have an infinite number of sides. By definition, a polygon is a closed figure with straight sides. In order for a shape to have straight sides, it must have a finite number of sides. If a shape had an infinite number of sides, it would form a circle rather than a polygon.
Conclusion
Polygons are closed shapes with straight sides and angles. The number of sides depends on the type of polygon, ranging from three in a triangle to infinity in a circle. Understanding the properties of polygons is crucial in various fields, including architecture, engineering, and mathematics.
With this knowledge, you can identify and classify different shapes and use them in practical applications. Keep exploring the fascinating world of polygons and their properties.
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Theorem 1:
If C is a point on the line AB, then
C = (1-t)A + tB.
Theorem 2:
If A, B, C are collinear, then real numbers x, y, z not all zero such that (bidirectional)
x+y+z=0 and xA + yB + zC = 0
Menelaus:
Best said: if a line cuts a triangle (you have to extend one of its sides), then the product of the ratio of each of the cuts = -1.
Ceva:
Best said: In a triangle, if lines are drawn from each of the edges to the opposite side and the lines are concurrent, then the product of the ratios of each of the cuts is 1. The centroid is an instance of Ceva, where each of the ratios in 1 because it's (1/2 / 1/2 ). Converse is also true. If product is 1, lines are concurrent.
Desargues (parallel case):
If the sides of two triangles are parallel, then the joins of the corresponding vertices are either concurrent or parallel. This point of concurrence is the center of perspective.
Desargues:
Two triangles, distinct vertices, if they're perspective with respect to a point, they're perspective with respect to a line.
Reciprocal figures:
Two triangles, point inside of one. If the three lines connecting that point to each vertex are parallel to the sides of the other triangle, then there exists a point in the other triangle that has the same property.
Dot (Inner) Product:
A . B is equal to |A| |B| cos(angle between). Which is essentially the product of the projection of one vector onto the other.
Theorem Pappus (parallel case):
Hexagon? If two pairs of opposite sides are parallel, then the third are also parallel.
Napoleon:
Draw equilateral triangles on the sides of any triangle, then the centers of those triangles create an equilateral triangle.
Other theorem:
If you have any triangle, draw squares on two of the sides and draw lines from the midpoints of these squares to the midpoint of the side you didn't draw a square off of. These two lines are equal length.
Wedge product (2D):
Length: |u| * |v| * sin (angle)
Geometrically: it's the area of the quadrilateral created by the two vectors.
Line is perp to uv plane.
Orientation Right hand rule.
Algebraic, u ^ v is: u1v2 - u2v1
u ^ u = 0
u ^ v + v ^ u = 0
ABC collinear <==> A^B + B^C + C^A = 0.
REVISIT TRIPLE PRODUCT OF 3 3D VECTORS.
ABC \in R^2 (and z=1): A B C = 0 <==> ABC collinear.
Where ABC \in R^2: 1/2(A B C) = area of triangle. Put z=1.
A B C = Volume of parallelepiped.
1/6 (A B C) = volume of pyramid with triangle base.
Algebraically:
w1 w2 w3
det(u1 u2 u3 ) = can do the diagonal lines + - thing.
v1 v2 v3
Law of Cosines:
Given an angle, theta, between b and a, c^2 = a^2 + b^2 - 2abcos(theta).
Median: line from vertex that intersects midpoint of opposite side. Intersection of medians is centroid.
Orthocenter: draw lines from vertex that make 90 degree with opposite side (altitudes), their intersection is the orthocenter.
Showing perpendicularity: just need to show dot product is zero. Really just use algebra.
Rotations:
If you know there's a certain angle between two vectors, then you can write one of the vectors as R(other). You can also write one vector in terms on another two, then do a rotation on that vector, and then split up the rotations to be rotations of the the two that made it. So, say you had R(u + v / 2), well that's = (R(u) + R(v))/2. And say you knew something about R(u) and R(v), then you've just made a little jump.
Barycentric Coordinates:
A, B, C three non-collinear points, any vector P may be expressed as P = xA + yB + zC, where x + y + z = 1.
Black hole problem: two lines and a point, all intersecting at a point, describe that point of intersection by using Desargues. Short version: create two triangles with these lines, show that they're perspective with respect to a line and therefore they are perspective with respect to a point. Essentially what you do is create one full triangle with your lines and then another side of a triangle. You show that these two have an axis of perspective.
Altitudes of triangles are concurrent proof: Draw the triangle, draw two of the altitudes and the point that they intersect at, and then you'll say "let's show that the line that goes through this intersection from the remaining vertex is actually an altitude". So you just need to show that some dot product is zero. Should be algebra.
Centroid proof: So like, for the centroid of a triangle being the average of the three sides, you connect the edges to the midpoint of their opposite side, and you notice that the intersection seems to divide these lines in a ration of 2:1, so you write the midpoint as two of the sides and then you're tasked with incorporating the last side, so you do so by considering this point on the line with a ratio of 2:1, and then you see that the algebra simplifies to sum of all the sides over three.
Menelaus Proof: Use Theorem 2.
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S^2:
Polar triangle of triangle x,y,z is defined as x'y'z' where x' = pole of great circle yz, on the same side as x. y', z' defined similarly.
T_xS^2 is tangent space to S^2 at vector x. v is unit tangent vector to segment xy if ||v|| = 1, and y=alpha*x + beta*v.
Length of Spherical Segment xy is arccos(x \dot y). Length = angle.
Side lengths of triangle abc are pi - angles of a'b'c'.
Law of Cosines: cos(c) = cos(a)cos(b) + sin(a)sin(b)cos(gamma).
Dual law of Cosines: cos(gamma) = -cos(alpha)cos(beta) + sin(alpha)sin(beta)cos(c). Note that the angle, gamma, depends on the opposite side, not just the other angles!
Law of sines: sin(a)/sin(alpha) = sin(b)/sin(beta) = sin(c)=sin(gamma)
Theta-lune: two intersecting great circles at angle theta, that area (for both sides) = 4*theta. Could think about 2pi case.
(alpha + beta + gamma) = pi + area
Sum of angles go between pi< and <3pi
H^2:
H^2 is all vectors u such that Phi(u,u)=-1.
Length of curve d(x,y) = arccosh(-\Phi(x,y))
phi: t—> xcosh(t) + usinh(t) where Phi(u,u)=1, phi stays in H^2 for all t. Also, d(x,s(t))=t.
Law of cosines: cosh(c) = cosh(a)cosh(b) - sinh(a)sinh(b)cos(gamma).
Dual law of cosines: cos(gamma) = -cos(alpha)cos(beta) + sin(alpha)sin(beta)cosh(c). Same thing as in S^2: gamma depends on opposite side, not just the other two angles.
Pi(d) = arcsin(1/ch(d)), where d is the vertical distance between a line and its limiting parallel. Pi defines the angle opposite the line.
The area of 3x ideal triangles is finite, and its area is pi.
(alpha + beta + gamma) = pi - area.
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AB is the diameter of a circle with centre O. C and D are two points on the circle on either side of AB, such that ∠CAB = 52° and ∠ABD = 47°. What is the difference (in degrees) between the measures of ∠CAD and ∠CBD?
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Lesson video
My name is Mr Kelsall and welcome to today's lesson about revision of angle properties.
And before we start, you'll need a pen and a piece of paper and somewhere quiet that you're not going to get disturbed.
Don't forget to remove any sort of distractions for example, put your mobile phone on silent or move it away completely.
And then pause the video and when you're ready, let's begin.
Our lesson today is all about revision of angle properties.
We will start by looking at angles on a line, angles around a point and angles on a right angle.
We'll then move on to angles within a triangle and quadrilateral.
We'll then look at parallel lines and solving problems within parallel lines and all the shapes.
And after that, it's quiz time.
As I've mentioned, you'll need a pencil and a piece of paper, ideally graph paper, and the star words for today we'll be talking about vertically opposite angles, right angles, we'll be looking at triangles which are scalene, isosceles and equilateral triangles.
Will be looking at quadrilaterals, mainly parallelograms, trapeziums, a rhombus, square, rectangle and we'll be using the word parallel.
In order to access this lesson you will need some basic angle facts and I'll go through these now.
You will need to understand that angles within a right angle add up to 90 degrees.
A right angle is notated by a small square.
If I split this and I say that this angle is 40 degrees can you account the remaining angle? So you do 90 degrees take away 40 degrees gives you 50 degrees which is the missing angle.
You can use this in a variety of ways.
Sometimes you won't always see right angles which look like a right angle, but if it's notated by a square it will be a right angle.
Second factor is you need to understand angles on a straight line add up to 180 degrees.
So that means if I look at my example here I know these angles add up to 180 degrees.
So 50 and 60 is 110.
180 takeaway 110 gives you 70 degrees or you could do it 110 and carry out adding up to 180 degrees to get 70 degrees.
But angles on a straight line add up to 180 degrees.
You'll also look vertically opposite angles.
And this is all to do with angles which are also on a straight line.
If I look at this example here, I can see that I've got 50 degrees as my first angle.
I'm trying to find my second angle which is on a straight line.
I'll do 180 degrees take away 50 degrees gives me 130 degrees.
I then need to find my other measure angle here.
Well, I've got another straight line there.
So 130 degrees add on something gives me 180.
So 180 take 130 gives me 50 degrees.
So I can prove that these angles are vertically opposite by using my angles on a straight line idea.
Our final angle is 130 degrees.
The next skill is to understand angles around a point.
So if I have a singular point I understand that all these angles add up to 360 degrees.
Sometimes I might need to use more than one step.
For example, this question I can say I've got a straight line there.
So 90 degrees, add 90 degrees, it gives me 180 degrees.
So I've found one missing angle there.
That means I've just got one angle remaining to find.
Next rule is the angles inside a triangle add up to 180 degrees.
Now you will get three style questions on this.
The first one is where you get a triangle with different angles and you'll be given these angles and you're asked to find a missing angle.
The second one will be understanding properties of an equilateral triangle where all three angles are equal and all three angles are 60 degrees.
Your third type of question is where you've got an isosceles triangle and you've got two sides which are the same length and therefore two angles which are the same size.
You might be given one angle and be asked to find the other two angles, or you might be given one of the two angles and asked to find the top angle.
But either way angles within a triangle add up to 180 degrees.
And your final skill is all to do with angles within a quadrilateral.
Now, normally with these style questions you will be either given three of the angles and asked to find a fourth angle.
So if I told you that was 90 degrees, 90 degrees and 60 degrees, you'll be asked to find the missing angle here.
So 90 add 90 is 180, add 60 is 240.
I need to add on another 120 to find my missing angle.
You might also get questions which ask you to use properties of the shape that you know.
So for example, this is a parallelogram and I know that opposite angles are equal size.
This is a rhombus and I know these angles are equal and these are angles are equal.
So you might be asked to use the properties of the shape that you given.
So our new learning and our first task today, you've got several questions that you can have a look and have a go on.
They are all to do with angles on a right angle, angles on a straight line, angles around a point, vertically opposite angles.
Before you do that there are three questions on the board and these are real thinking questions.
The first one says, the angles on the right angle add up to 90 degrees, prove this.
How can you prove that angles on a right angle add up to 90 degrees? The next one is angle on a straight line add up to 180 degrees.
Again, prove this, convince me, show me.
Imagine I'd never seen an angle before and I need to understand that angles on a straight line add up to 180 degrees, prove it to me.
And the same idea with 360 degrees.
Pause the video, and then when you're ready, press play to continue.
So perhaps the easiest way to prove that angles on a right angle add up to 90 degrees is by finding yourself an angle measure.
So you can fold a piece of paper.
Fold it again, it will give you a 90 degree right angle.
You can measure this on a protractor.
To prove that angles in a right angle add up to 90 degrees you can take this piece of paper, you could fold it in half and you can measure each side to prove that it add up to 90 degrees.
This is what we call your base fact.
And the reason it's your base factor is because you can use it to prove the other facts.
My second thing is to prove that angles on a straight line add up to 180 degrees.
Well, mostly much now I've got my right angle there.
If I continue this and I do another right angle that gives me a straight line.
So I've proved that angles on a straight line add up to 180 degrees.
It doesn't have to be a straight line that we normally see.
You might see a straight line but it's at an angle and I might draw it like this.
I might have a right angle and another right angle.
But I know that two right angles, 90 and 90 add up to 180 degrees.
So that line there, must be 180 degrees.
And my final point I need to prove is 360 degrees.
If I know that a straight line is 180 degrees, then I know another straight line is also 180 degrees and 180 degrees and 180 degrees is 360 degrees.
This brings us to using these facts to start by solving problems. Well, the first question asks angles in a right angle add up to 90 degrees.
I know I've got used 70 degrees of the 90 which means that this is a further 20 degrees.
question.
Question number two relies on vertical opposite angles to understand that that is 50 degrees.
You can work out by sum of that one straight line.
So this angle must be 130 degrees and this is another straight line.
So 130 add 50 is 180.
You can do with the same with the straight line at the bottom.
50 add something equals 180 must be 130 degrees.
And your final task, you're given two angles out of three possible angles.
So my three angles add up to 180 degrees.
One is 50, one is 70 together those add up to 120 degrees which means that add a further 60 degrees and that gives you 180 degrees.
So for our final two questions, the first one is looking at angles around a point.
Now I know that these lines here look like they're right angles.
And normally, if you aren't told they different you could presume they're right angles.
However, I've been told that this angle is 85 degrees.
So that means I can't presume that either of these are right angles.
So I need to use my angles on straight line properties to find out this missing angle.
85 degrees and something equals 180 degrees.
I can think, well, I'm going to start with my 180 degrees.
I'll take away 80 to give me 100 and I'll take away another five which means that this missing angle is 95 degrees.
Now, I've got some information.
I've got three angles out of four angles.
I know all four angles add up to 360 degrees.
So I need to find out what these three angles add up to.
I'll start with the biggest one, 135.
I'm going to add on the 95.
To do that I'll add on 100 and take away 5.
So that's 235 take away 5 that's, 230.
And then going to add on 85.
So 230, when I add on 100, which is 330 and I'll take away 15, which is 315 degrees.
So I know that these three angles add up to 315 degrees.
So all that's left for me to do is say, well I need to find the difference between 360 degrees and 350 degrees.
You can count from 315, add 5 is 320, add 40 gives you 360.
Or you could use a written method.
You could say 360 take away 315.
So 10 take five is five.
Five take one is four, three take three is zero.
So this missing angle is 45 degrees.
And my final one involves a little bit of thinking and I need to use a few different angle properties.
The starting point for me is that I have a right angle here and it's been split into three equal parts.
So 90 degrees divided by three gives me 30 degrees.
So I know this angle, this angle and this angle are 30 degrees each.
Oh, I only need some bar angle is 30 degrees.
And I need to find out what this missing angle here is.
I know angles around the point add up to 360 degrees.
And I know that I've used 30 degrees of this, which means that that big angle the reflex angle is 330 degrees.
And that brings us to the develop learning for today.
So here we have two tasks and then several questions.
Task Number one, look at the angles in the triangle, they add up to 180 degrees.
Can you approve this? Same with angles in a quadrilateral they add up to 360 degrees.
Can you approve this too? Once you've done that, have a go at the questions on the screen.
Pause the video and press play when you're ready to continue.
So I'm trying to prove that triangle has angles which add up to 180 degrees.
So I have my triangle here and if I rip off this corner and put it here.
If I rip off this corner and rotate it and put it here and then if I rip off this corner, turn it around and put it there.
What I can see is all three of these angles combined make a straight line.
Now I know the angles on a straight line add up to 180 degrees, therefore angles inside a triangle must add up to 180 degrees.
I can apply a similar knowledge to angles inside of quadrilateral.
If I take myself a quadrilateral, I start from any vertex and I join a vertex to all the other vertices.
I can see that I create two triangles.
Now I know one triangle has angles of 180 degrees, and I know the other one has angles of adding up to 180 degrees.
Therefore, I know the angles inside this quadrilateral must add up to 360 degrees.
And I can repeat this with a range of different quadrilaterals.
So if I try a trapezium, I choose a corner and I join up that corner to all the other corners, I create two triangles.
I can do the same thing.
I even choose unusual shapes that we don't normally think of when we looking at quadrilaterals.
And I'm going vertically one vertex to the vertices, I create two triangles.
Therefore I prove that angles inside a quadrilateral add up to 360 degrees.
The next we need to look at these style of questions.
Now my first question, I'm given two angles and I need to find the third angle.
50 degrees add 30 degrees is 80 degrees.
So my third and missing angle must be 100 degrees.
Be careful of this, because these triangles deliberately drawn incorrectly.
Because I know that this angle is less than a right angle so it should be an acute angle.
So it should be less than 90 degrees but actually using the other numbers, it comes out as 100 degrees.
Be aware of things like this.
My second triangle is the green one and I can see that these lines here tell me all three sides of equal length.
If all three sides are equal length, all three angles are equal size.
So I'm thinking 180, split it between the three angles.
I can do this mentally because I can do 18 divided by three is six, therefore 180 divided by three is 60.
So each one of these angles is 60 degrees.
And finally, my third triangle is an isosceles triangle.
I know that it has two base angles that are equal and I've been given one angle which is 50 degrees.
180 degrees take away 50 degrees, I can do 18 take five is 13.
So 180 take 50 is 130.
So I know that these two angles add up to 130 degrees.
Therefore all I need to do is split 130 degrees by two.
You could do it mentally and do 13 divided by two is 6.
5.
Therefore 130 divided by two is 65.
Or you could use a written method like I'm going to use here.
How many twos go into one, zero carry the one.
How many twos going to 13, six, it's one leftover.
How many twos into 10, five.
You can solve it whichever way you choose but each of these angles is 65 degrees.
Then we get to some more complicated triangles but I can still use the properties of the triangles that I know.
If I start, I know I have each of these are 60 degree angles.
I know I've been given a 40 degree angle there and I know angles around the point here, add up to 360 degrees.
I'm just going to draw this bit here.
So I've been given 60, 40, that adds up to 100 degrees.
Therefore, my remaining two angles must add up to 260 degrees.
Because 100 degrees add to 260 degrees is 360 degrees.
Therefore if I divide 260 by two, I'll find that each of these angles is 130 degrees.
I then need to come back and consider my base angles here.
I know my top angle is 40 degrees.
So 180, take 40 leaves 140, then split in 140 by two.
I could split 14 by two is seven.
So 140 by two is 70.
So each of these angles is 70 degrees.
Now leaves me my final angle.
Now my starting point here is that I know from 12 to 3 o'clock I form a right angle.
The next thing I know is this right angle has been split equally into one, two, three parts.
Therefore 90 degrees divided by three is 30 degrees.
So this angle here must be 30 degrees.
It's going to clean the screen.
If I know this angle is 30 degrees, I need to account this angle.
Well, 360 takeaway 30 is 330.
So this reflex angle here must be 330 degrees.
And I also be asked, what are these angles here? Well, I know this is an isosceles triangle because the distance from the centre to one is the same size as the distance between the centre to two.
So 180 degrees take away 30 degrees leaves me 150 degrees.
150 divided by two means that these angles are both 75 degrees.
Now it brings us to our independent task for today.
Before we start this question I have to tell you that all the questions so far involve you really using all the angle laws that you have so far and really thinking through what you know as well as what you don't know.
This question is exactly the same.
It uses all the properties we've discussed throughout the lesson, though you have to really think them through to understand which angles are which and how you can find the missing angles.
Pause the video and when you're ready, press play to continue.
My starting point is this angle here.
I know it's a right angle.
I know if I split it between one and two, I know that each of these degrees so from 12 to 1 is 30 degrees.
From 1 to 2, 30 degrees, 2 to 3, 30 degrees.
This angle here is 30 degrees.
Next thing I know this angle covers two hours.
It goes from 10 to 11, 11 to 12.
So I know that must be 60 degrees.
I could use the properties of angles on a straight line to find out this angle, and I know I've used 60 degrees of my 180 degrees so this must be 120 degrees.
And now I have a right angle here and a right angle here.
In fact, I'm going to add in all my right angles cause that will help me.
You'll notice there's quite a lot of right angles.
Okay so that eliminates a lot of angles we need to find.
I'm now going to to look at this angle.
If you look closely, this forms a quadrilateral, it's actually a trapezium, isn't it? Because I've got one pair of parallel sides and I know I've got an angle of 90 degrees, 90 degrees which is 180 and then a further 120 degrees which if I add 20 gives me 200 and 100 gives me 300.
So that means this missing angle must be 60 degrees.
I'm now going to move on to the triangle in the corner here.
Now I've got one, two angles left to find.
I'm going to clear this and show you how I found these angles.
You remember parallel lines.
If you have a line dissecting them, you find out this angle, I also have this angle, this angle and this angle.
Oh, have a look here.
I have parallel lines and if I extend this line, if I know this angle I can find out this angle and this angle, I can extend this a bit more, this angle too.
Perhaps I know that this angle is 30 degrees because I know that from between 12 to one o'clock is 30 degrees.
Therefore this must be 30 degrees, 30 degrees, 30 degrees.
If I go back to this triangle, I've got a 30 degree angle, a 90 degree angle, which takes me 120 degrees.
That means this final missing angle here must be 60 degrees.
I was going to say, you might have found this differently.
You might have said that from one o'clock to three o'clock covers two gaps.
So that is 60 degrees.
Then you've got your 90 degrees, 60 degrees and 90 degrees is 150, which means that is 30 degrees.
There are a range of ways that you can solve the missing angles but it's just a thought process of understanding what rules you can apply to this problem.
Congratulations on completing your task.
If you'd like to please ask your parent or carer to share you work on Twitter tagging @OakNational and also #LearnwithOak.
Now before we go, please complete the quiz.
And so that brings us to the end of today's lesson on revision of angle properties.
A really big well-done for all the fantastic learning that you've achieved.
Now, before you finish perhaps we'll quickly review your notes and try to identify the most important part of your learning from today.
Well, all that's left for me to say is thank you, take care and enjoy the rest of your learning for today.
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Elements of Geometry
From inside the book
Results 6-10 of 82
Page 23 ... Hence the greatest straight line that can be inscribed in a circle is equal to its diameter . THEOREM . 101. A straight line cannot meet the circumference of a circle in more than two points . Demonstration . If it could meet it in ...
Page 24 ... hence the angle ACD = EOG . Now , if the semicircle ADB be placed upon EGF , because the angle ACD = EOG , it is evident , that the radius CD will fall upon the radius OG , and the point D upon G , therefore the arc AMD is equal to the ...
Page 26 ... hence the three distances OA , OB , OC , are equal ; therefore the circumference , described from the centre O with the radius OB , will pass through the three points A , B , C. It is thus proved , that the circumference of a circle may ...
Page 29 ... Hence , if two circles touch each other , either internally or externally , the centres and the point of contact are in the same straight line . 118. Scholium . All the circles , which have their centres in the straight line CD and ...
Page 32 ... hence two sectors ACB , ACD , taken in the same circle or in equal circles , are to each other , as the arcs AB , AD , the bases of these sectors . It will be perceived therefore , that the arcs of a circle , which are used as a measure ...
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So, the coordinates of midpoint of side $PQ$ is given as $D\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$ .
The coordinates of midpoint of side $QR$ is given as $E\left( \dfrac{{{x}_{2}}+{{x}_{3}}}{2},\dfrac{{{y}_{2}}+{{y}_{3}}}{2} \right)$ .
And the coordinates of midpoint of side $PR$ is given as $F\left( \dfrac{{{x}_{1}}+{{x}_{3}}}{2},\dfrac{{{y}_{1}}+{{y}_{3}}}{2} \right)$ .
Now let's find the equation of line perpendicular to the sides and passing through the midpoint of the sides i.e. perpendicular bisector of sides. So, the equation of perpendicular bisector of $PQ$ is given as:
Note: While simplifying the equations , please make sure that sign mistakes do not occur. These mistakes are very common and can cause confusions while solving. Ultimately the answer becomes wrong. So, sign conventions should be carefully taken.
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Second place goes to the "teensy radius" corner, followed by the perfectly-square corners. For what it's worth, the park-and-burn effect largely lands at the outside of the corners, so the effect is somewhat mitigated without over-extending the line segments.
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equilateral triangle area worksheet
Triangle Area Worksheet – Triangles are one of the most basic shapes found in geometry. Understanding triangles is crucial to learning more advanced geometric concepts. In this blog, we will cover the different kinds of triangles including triangle angles and the methods to calculate the size and perimeter of a triangle and will provide instances of each. Types of Triangles There are three kinds of triangles: equilateral, isosceles, as well as scalene. Equilateral triangles consist of … Read more
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The First Six Books: Together with the Eleventh and Twelfth
to the other angles b, each to each, to which the equal fides are oppofite; therefore the angle ACB is equal to the angle CBD; and because the ftraight line BC meets the two ftraight lines b 4. I. AC, BD, and makes the alternate angles ACB, CBD equal to one another, AC is parallel to BD; and it was shown to be equal to it. Therefore ftraight lines, &c. Q. E. D.
TH
HE oppofite fides and angles of parallelograms are equal to one another, and the diameter bifects them, that is, divides them in two equal parts.
N. B. A parallelogram is a four fided figure, of which the oppofite fides are parallel; and the diameter is the ftraight line joining two of its oppofite angles.
Let ACDB be a parallelogram, of which BC is a diameter; the oppofite fides and angles of the figure are equal to one an other; and the diameter BC bifects it.
C
B
D
c 27. I.
a 29. I.
Because AB is parallel to CD, A and BC meets them, the alternate angles ABC, BCD are equal to one another; and becaufe AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD are equal a to one another; wherefore the two triangles ABC, CBD nave two angles ABC, BCA in one, equal to two angles BCD, CBD in the other, each to each, and one fide BC common to the two triangles, which is adjacent to their equal angles; therefore their other fides fhall be equal, each to each, and the third angle of the one to the third angle of the other b, viz. b 26. s. the fide AB to the fide CD, and AC to BD, and the angle BAC equal to the angle BDC: And because the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB, the whole angle ABD is equal to the whole angle ACD: And the angle BAC has been fhown to be equal to the angle BDC; therefore the oppofite fides and angles of parallelograms are equal to one another; alfo, their diameter bifects them; for AB being equal to CD, and BC common, the two AB, BC are equal to the two DC, CB, each to each; and the angle ABC
Book I. is equal to the angle BCD; therefore the triangle ABC is equal to the triangle BCD, and the diameter BC divides the parallelogram ACDB into two equal parts. Q. E. D.
C 4. I.
See N.
See the 2d
ARALLELOGRAMS upon the fame bafe and between the fame parallels, are equal to one another,
PAR
Let the parallelograms ABCD, FBCF be upon the fame
and 3d fi bafe BC, and between the fame parallels AF, BC; the parallelogram ABCD fhall be equal to the parallelogram EBCF.
gures.
If the fides AD, DF of the pa- rallelograms ABCD, DBCF oppoûte A to the bafe BC be terminated in the fame point D; it is plain that each of the parallelograms is double a of the triangle BDC; and they are there- fore equal to one another.
But, if the fides AD, EF, oppofite B to the bafe BC of the parallelograms
ABCD, EBCF, be not terminated in the fame point; then, becaufe ABCD is a parallelogram, AD is equal a to BC, for the fame reafon EF is equal to BC; wherefore AD is equal b to EF; and DE is common; therefore the whole, or the remainder, AE is equal to the whole, or the remainder DF; AB alfo is equal to DC; and the two EA, AB are therefore equal to
A
DE
FAE
DF
WW
B
C
B
C
the two FD, DC, each to each; and the exterior argle FDC is equal to the interior EAB; therefore the bafe EB is equal to the bafe FC, and the triangle EAB equal to the triangle FDC; take the triangle FDC from the trapezium ABCF, and from the fame trapezium take the triangle EAB; the remainders therefore are equal, that is, the parallelogram ABCD is equal to the parallelogram EBCF. Therefore parallelograms upon the fame bafe, &c. Q. E. D.
PROP
ARALLELOGRAMS upon equal bases, and between the fame parallels, are equal to one another.
PA
B
C F
G
Join BE, CH; andbecause BC is equal to FG, and FG to a EH, BC is equal to EH; and they are pa- a 34. X. rallels, and joined towards the fame parts by the ftraight lines BE, CH: But ftraight lines which join equal and parallel ftraight lines towards the fame parts, are themselves equal and parallel; therefore EB, CH are both equal and parallel, and b 33. I. EBCH is a parallelogram; and it is equal to ABCD, because c 35. 1. it is upon the fame base BC, and between the fame parallels BC, AD: For the like reafon, the parallelogram EFGH is equal to the fame EBCH: Therefore alfo the parallelogram ABCD is equal to EFGH. Wherefore parallelograms, &c. Q. E. D.
TR
PRO P. XXXVII.
с
THEOR.
RIANGLES upon the fame bafe, and between the fame parallels, are equal to one another.
E
A
D
F
Let the triangles ABC, DBC be upon the fame base BC and between the fame parallels AD, BC: The triangle ABC is equal to the triangle DBC.
Produce AD both ways to the points E, F, and thro' Burawa BE para lel to CA; and thro' C draw CF paral
lel to BD: Therefore each
B
C
2 31. I.
of the figures EBCA, DBCF is a parallelogram; and EBCA is equal to DBCF, because they are upon the fame base BC, and b 35. I. between the fame parallels BC, EF; and the triangle ABC is
the
Book I.
€ 34. I.
the half of the parallelogram EBCA, because the diameter AB bifects c it; and the triangle DBC is the half of the parallelogram DBCF, because the diameter DC bifects it: But the d 7. Ax. halves of equal things are equal d; therefore the triangle ABC is equal to the triangle DBC. Wherefore triangles, &c. Q. E. D.
TRI
RIANGLES upon equal bafes, and between the fame parallels, are equal to one another.
Let the triangles ABC, DEF be upon equal bafes BC, EF, and between the fame parallels BF, AD: The triangle ABC is equal to the triangle DEF.
Produce AD both ways to the points G, H, and through B draw BG parallel a to CA, and through F draw FH parallel to
ED: Then each of G
the figures GBCA, DEFH is a paralle logram; and they are equal to one an- other, because they are upon equal bafes BC, EF, and between
the fame parallels
B
C 34. T.
d 7. Ax.
a 31. I.
BF, GH; and the triangle ABC is the half of the parallelogram GBCA, because the diameter AB bifects it; and the triangle DEF is the half of the parallelogram DEFH, because the diameter DF bifects it: But the halves of equal things are equal; therefore the triangle ABC is equal to the triangle DEF; Wherefore triangles, &c. Q. E. D.
E
QUAL triangles upon the fame bafe, and upon the fame fide of it, are between the fame parallels.
Let the equal triangles ABC, DBC be upon the fame bafe BC, and upon the fame fide of it; they are between the fame parallels.
Join AD; AD is parallel to BC; for, if it is not, through the point A draw AE parallel to BC, and join EC:
The tri
angle
b
A
E
b 37. T
angle ABC is equal to the triangle EBC, because it is upon Book I the fame base BC, and between the fame parallels BC, AE: But the triangle ABC is equal to the triangle BDC; therefore also the triangle BDC is equal to the triangle EBC, the greater to the leis, which is impoffible: There, fore AE is not parallel to BC. In the fame manner, it can be demonstrated that no other line but AD is parallel to
EF, in the fame ftraight line BF, and towards the fame parts; they are be tween the fame parallels.
Join AD; AD is parallel to BC: For, if it is not, through A draw AG parallel to BF, and join GF:
B
Fb 38. I
CE
The triangle ABC is equal to the triangle GEF, because they are upon equal bafes BC, EF, and between the fame parallels BF, AG But the triangle ABC is equal to the triangle DEF; therefore alfo the triangle DEF is equal to the triangle GEF, the greater to the lefs, which is impoffible: Therefore AG is not parallel to BF: And in the fame manner it can be demonftrated that there is no other paral- lel to it but AD; AD is therefore parallel to BF. Wherefore equal triangles, &c. Q. E. D.
IF a parallelogram and triangle be upon the fame base, and between the fame parallels; the parallelogram shall be double of the triangle.
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The history of geometry
Early on geometry goes back to about 3000 BC, as one of the 1st advancements of geometry. That started in The european union as the Egyptians tried it in many ways such as, surveying of land, structure of the pyramids and astronomy. The next advancement came from the Babylonians in 2000-500 BC. Ancient clay-based tablets showed that the Babylonians knew regarding the pythagorean relationships.
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In 750-250 BC the Greeks, utilized experimental geometry like Egypt and Babylonia had. They will created the 1st formal mathematics of all kinds by arranging geometry with rules of logic. The next advancement in geometry video game from Euclid. In 300 BC he wrote a text titled Elements. This kind of stated that ideas could possibly be proven by using a small group of statements (postulates).
The five postulates were:
A straight line part can be driven joining virtually any two points.
Given any kind of straight range segment, a circle may be drawn having the segment as being a radius and one endpoint as the middle.
Perfectly angles happen to be congruent.
If two lives are drawn which intersect a third range in such a way that the sum in the inner aspects, then the two lines without doubt, must meet each other quietly if expanded indefinitely. There were a lot of controversy above the fifth postulate.
The fifth postulate states Provided a range and an area not on the line, it is possible to draw accurately one line through the given stage parallel towards the line. Many mathematicians through the next hundreds of years unsuccessfully attempted to prove this postulate.
In 1600 AD, Rene Descartes manufactured one of the greatest developments in angles. He linked algebra and geometry. A myth is the fact he was watching a fly on the ceiling when he conceptualized of finding points on a plane with a pair of figures. Fermat likewise discovered put together geometry but Descartes' type is the 1 we make use of today.
In the early 1800's Bolyai and Lobachevsky the initial non-euclidean geometries. Since no person could prove Euclids fifth postulate, they devised new geometries with odd notions of parallelism. Back in the 1800's to early 1900's, Gauss and Riemann set the foundation intended for differential geometry. Differential geometry combines geometry with the techniques of calculus to provide a way for studying angles on rounded surfaces.
Also back in the 1800's to early 1900's Mandelbrot and a few other experts researched fractal geometry. Fractals are geometric figures that model a large number of natural set ups. The invention of computers provides helped with study regarding fractals.
This is an essay portal with lists of topics, sample essays and research papers, and writing tips for high school and college students. Best ideas for your writing assignments. Our goal is to help students write essays and other papers.
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For those who don't understand what the above formula does, it basically draws a line behind the player and then determines which side of that line the target is on. However, the above formula made sense to me because I am able to think of it in cartesian coordinates and not polar ones.
I'm trying to think of a way to determine the shortest way to match the angle of another object, and I can't seem to find a cartesian way to think of it.
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Hint: In order to solve the given question , we will be using the theorem which states that the angle subtended by an arc at the center is double the angle subtended by it on any point on the remaining part of the circle .
Complete step-by-step answer:
Let us consider the angle ANB first , we can see in the image , AB is the diameter . Therefore, the angle subtended by the arc AB at the centre is ${180^ \circ }$. Give the centre a name like X. So, $\angle AXB = {180^ \circ }$ . Using the theorem which states that the angle subtended by an arc at the center is double the angle subtended by it on any point on the remaining part of the circle . $\angle AXB = 2\angle ANB$ $\angle ANB$is the angle made at any point on the circle . \[ \angle AXB = 2\angle ANB \\ \dfrac{{\angle AXB}}{2} = \angle ANB \\ \dfrac{{{{180}^ \circ }}}{2} = \angle ANB \\ \angle ANB = {90^ \circ } \; \] Similarly , it can be said that angle inscribed in a semicircle is a right angle so Angle ANB = Angle AOB = Angle AMB=${90^ \circ }$, because like \[\angle ANB = {90^ \circ }\],$\angle AOB,\angle AMB$are also the angles made at any point on the circle . So, the correct answer is "\[{90^ \circ }\]".
Note: An arc of a circle is any portion of the circumference of a circle. To recall, the circumference of a circle is the perimeter or distance around a circle. Therefore, we can say that the circumference of a circle is the full arc of the circle itself.
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1d Vs 2d Vs 3d Vs 4d
Understanding Dimensions: 1D vs 2D vs 3D vs 4D
In the realm of mathematics and physics, dimensions play a crucial role in understanding the fabric of our universe. From simple lines to complex structures, dimensions help us describe and analyze the world around us. In this article, we'll delve into the differences between 1D, 2D, 3D, and 4D, exploring their definitions, characteristics, and real-world applications.
1D (One Dimension)
A one-dimensional object is a straight line that extends infinitely in two directions. It has no width, height, or depth, and can be represented by a single coordinate (x).
Characteristics:
Length: The only dimension that exists in 1D.
No width or height: A 1D object has no width or height, making it a single point in space.
Real-world applications:
Music: Music can be thought of as a 1D concept, as it consists of a sequence of notes that flow in a linear fashion.
Time: Time can be considered 1D, as it moves in a straight line from past to present to future.
2D (Two Dimension)
A two-dimensional object is a flat surface that has both length and width, but no depth. It can be represented by two coordinates (x, y).
Characteristics:
Length and width: 2D objects have both length and width, but no depth.
Flat surface: 2D objects are flat and do not have thickness.
Real-world applications:
Images: Photographs and images are 2D representations of the world.
Maps: Maps are 2D representations of geographical areas.
3D (Three Dimension)
A three-dimensional object is a solid object that has length, width, and depth. It can be represented by three coordinates (x, y, z).
Characteristics:
Length, width, and depth: 3D objects have all three dimensions.
Volume: 3D objects occupy space and have volume.
Real-world applications:
** Objects:** Everyday objects, such as chairs, tables, and buildings, are 3D.
Design: 3D modeling is used in various industries, including architecture, engineering, and animation.
4D (Four Dimension)
A four-dimensional object is a theoretical concept that adds an additional dimension to the three dimensions we are familiar with. This additional dimension is often referred to as time.
Characteristics:
Length, width, depth, and time: 4D objects have all four dimensions.
Time as a dimension: 4D objects exist in a space-time continuum.
Real-world applications:
Theoretical physics: 4D is used to describe the universe in theories such as Einstein's theory of relativity.
Computer graphics: 4D is used in computer graphics to create animations and simulations.
In conclusion, each dimension builds upon the previous one, adding more complexity and depth to our understanding of the world. From the simplicity of 1D lines to the intricacies of 4D space-time, dimensions help us navigate and make sense of our surroundings.
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Cross Products
The cross product is an essential concept in vector mathematics, especially in physics and engineering. It involves the multiplication of two vectors to produce a new vector that is perpendicular to both input vectors. The cross product is typically used to calculate the area of a parallelogram, the torque applied to an object, and the orientation of magnetic and electric fields.
How to calculate the cross product
To calculate the cross product of two vectors A and B in three-dimensional space, we use the following formula:
Flashcards covering the Cross Products
Practice tests covering the Cross Products
Get more help with the cross product
A tutor can equip your student with plenty of math tricks, including working with vectors and much more. Varsity Tutors carefully vets each tutor, ensuring that your student always works alongside someone who truly understands these concepts. If your student isn''t clear about cross-multiplying and other important concepts covered in class, they can ask as many questions as they like during these 1-on-1 sessions. Reach out today, and Varsity Tutors will match your student with a suitable
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Trilateration
This online calculator solves the problem of trilateration - determining the coordinates of a point by the distance from that point to three other points with known coordinates.
The result of solving the problem can be one of three:
There is no point the distances from which to the other three correspond to the given ones
There is exactly one point, the distances from which to the other three correspond to the given ones.
There are two points whose distances to the other three correspond to the given ones
Calculation formulas and illustrations for each case are given below the calculator.
Trilateration
Coordinates of the first known point (x, y, z)
Distance to the first point
Coordinates of the second known point (x, y, z)
Distance to the second point
Coordinates of the third known point (x, y, z)
Distance to the third point
Result
Solutions
Calculation precision
Digits after the decimal point: 2
Trilateration
The locus of points lying at a given distance r from a point with coordinates (x, y, z) is the surface of a sphere of radius r with the center at point (x, y, z). Thus, from the point of geometry, the problem of trilateration is to find the coordinates of the intersection of three spheres. These coordinates are found by solving a system of equations based on the following reasoning1:
Each pair of spheres intersect in a circle. Its center is on a straight line connecting the centers of the spheres. The circle itself lies in a plane perpendicular to that line. Consider three spheres: sphere 1 with radius r1 centered at the point O1(x1, y1 >, z1), sphere 2 with radius r2 centered at point O2(x2, y2, z2) and sphere 3 with radius r3 centered at O3 >(x3, y3, z3).
The equation of the plane in which the circle formed by the intersection of spheres 1 and 2 lies is as follows:
The equation of the plane in which the circle formed by the intersection of spheres 1 and 3 lies is as follows:
The equation of the plane of the triangle formed by the centers of the spheres:
The intersection of the first two planes gives a line perpendicular to the last plane. The intersection of this line with the plane of the triangle is the perpendicular from the desired point of intersection of the spheres to the plane of the triangle formed by the centers of the spheres. This intersection point belongs to all three planes, and its coordinates are the solution of the system of three linear algebraic equations given above.
Having solved this system, we obtain the coordinates of the point O (x0, y0, z0).
Then the coordinates of the intersection point of the three spheres are determined by the following formulas:
The expression for calculating k determines the number of solutions. If the difference under the square root is positive, then we have two solutions (+k, -k) - two intersection points of the three spheres, illustrated by the picture below
Example of two intersections
If the difference under the root is zero, then we have exactly one point of intersection of two spheres:
Example of one intersection
Finally, if the expression under the root is negative, then there are no solutions - the spheres do not intersect
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3. Warm Up Activity On a piece of paper, identify the following angles.
4. Warm Up Activity Answers Review the questions from the previous slide. Acute Right Obtuse Acute
5. Instruction: Types of Angles Read through the vocabulary and review the diagrams. Then watch the video linked here. 1) Interior Angle: an angle inside a shape 2) Exterior Angle: is an angle outside of a shape, made by the side of a shape and a line drawn out from an adjacent side 3) Acute Angle: an angle that's less than 90 degrees 4) Obtuse Angle: an angle that is greater than 90 5) Right Angle: a 90 degree angle 6) Complementary Angle: two angles that when added together equal 90 degrees Today we are 7) Supplementary Angle: two angles that when added mainly focusing on Interior and together equal 180 degree Exterior Angles!
6. Practice: Types of Angles On a piece of paper: Identify the exterior and interior angles shown in the problems. 1) 2) 3) 4)
7. Practice: Types of Angles ANSWERS On a piece of paper: Identify the exterior and interior angles shown in the 1) 2) 3) 4)
8. Instruction: Triangle Sum Theorem Read through the theorem and steps. Watch the provided video here. Then review the examples on the next slide. The triangle sum property states that the sum of the three interior (inside the triangle) angles in a triangle is always 180 degrees. 1) Write an equation that adds all three angle 2) Set the equation equal to 180 degrees. 3) Solve for the variable. 4) Plug the value of the variable (the answer) back into any Don't forget your steps angle expression that you need to find the value of. for solving equations!
12. Instruction: Exterior Angle Theorem Read through the theorem and steps. Watch the provided video here. Then review the examples on the next slide. The measure of an exterior angle of a triangle is equal to the sum of the measures of the two remote interior angles. What if you are not given all the angles? 1) Write an equation: remote interior angle + remote interior angle = exterior angle 2) Solve for the variable. 3) Plug the value of the variable (the answer) back into any angle expression that you need to find the value of.
13. Instruction: Exterior Angle Theorem Review the examples.
14. Practice: Exterior Angle Theorem On a piece of paper: Solve for the variable, then find the missing angle(s). 3.
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The orthocenter of a triangle is the point where all the altitudes of the triangle intersect. To find the orthocenter of the triangle with vertices at (4, 3), (7, 4), and (2, 8), you would need to find the intersection point of the altitudes.
The altitudes of a triangle are perpendicular lines drawn from each vertex to the opposite side. You can find the equations of these lines and then solve for their intersection point.
Once you have the equations of the altitudes and their intersection point, you will have found the orthocenter of the triangle
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Hint: Problems like these are quite simple and don't require much knowledge of mathematics. However we need to know what a vertical plane and a horizontal plane mean. We need to explore a bit in the open world and in our locality and environment to be able to find the answer to these types of problems. We can define a horizontal plane as any plane or surface which is parallel to the surface of the earth, or in other words we can also say that the plane which is at an angle of zero degree is known as the horizontal plane. Any plane which is perpendicular to the horizontal plane is known as the vertical plane.
Complete step by step answer: Now we start off with the solution to the given problem by trying to look around ourselves and find instances which represent the horizontal plane and vertical plane. We search for a plane which is parallel to the ground, and we can easily find one as the floor or ceiling of our room. The floor or ceiling is a perfect example of a horizontal plane or horizontal surface. Perpendicular to the floor, the wall is a great example of a vertical plane.
Note: These questions require some practical knowledge of the surroundings rather than any theoretical knowledge of mathematics. However we must have a clear cut understanding of what a horizontal and a vertical plane is. In theoretical perspective, to find a plane perpendicular to another plane, we need to have the equation of the horizontal plane as well as the direction ratios of the vertical plane.
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Over 2,000 years ago, a Greek philosopher called Pythagoras created a very famous theorem about triangles. It lets you work out the length of any side in a right-angled triangle!
What is the Pythagorean theorem?
The Pythagorean theorem states:
"In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the two shorter sides."
This may sound a bit complicated, but it makes a lot more sense when we look at what it's saying in the form of a picture:
When squares are drawn along each side in a right triangle, the total area of the two smaller squares (those along the shortest sides, a and b) is the same as the area of the largest square (the square along the longest side, c).
By working out the area of each square, we can find the length of any side in a right triangle. And luckily, Pythagoras created a handy formula to help us do this!Pythagorean theorem formula
In this equation, "C" represents the longest side of a right triangle, called the hypotenuse. "A" and "B" represent the other two sides of the triangle.
To use the Pythagorean theorem formula, we need to know the length of any two sides in a right triangle. We can then rearrange the formula to find the side we're looking for.
So, if we take the formula:
a² + b² = c²
We can rearrange it to help us find the length we're missing:
To find the length of Side A: a² = c² – b²
To find the length of Side B: b² = c² – a²
To find the length of Side C: c² = a² + b²
This may sound complicated, but don't worry! It's easy to use once you know how. Let's take a look at some examples.
How to use the Pythagorean theorem
Now that we know the Pythagorean theorem formula is, let's talk about what we can use it for. The Pythagorean theorem can be used to solve for the hypotenuse or the shorter sides of a right triangle.
Finding the hypotenuse of a triangle
If we don't know the length of the hypotenuse of a right triangle (aka the longest side), we can work it out using Pythagoras' theorem. The hypotenuse is represented by c in the Pythagorean theorem formula: a² + b² = c². By plugging in the given values of Side A and Side B, we can solve for the hypotenuse — Side C!
Let's try an example!
Looking at the picture below, solve for Side C.
To start, we can flip the Pythagorean theorem equation to make it easier to read:
a² + b² = c² becomes c² = a² + b²
Next, plug the values of Side A and Side B into the equation since they are given. Side A = 4cm and Side B = 3cm. This gives us:
c² = 4² + 3²
Square Sides A and B to get:
c² = 16 + 9
Add 16 and 9 solve for a²:
c² = 25
Now we're in the final stretch. We just need to find the square root of 25 to find the length of the hypotenuse (Side C)!
c² = √25
Therefore, the length of Side C = 5cm.
Finding the short side of a right triangle
We can also use Pythagoras' theorem when we don't know the length of one of the shorter sides of a right triangle.
We can rearrange the formula to help us find the side we don't know. For example:
If we don't know Side A: a² = c² – b²
If we don't know Side B: b² = c² – a²
Let's practice! In the example below, solve for Side A.
To work out the length of Side A, start by rearranging the formula to become:
a² = c² – b²
Next, plug in the values of Side C and Side B. (Remember: Side C is the hypotenuse!)
a² = 5² – 3²
Square Sides B and C to get:
a² = 25 – 9
Subtract 9 from 25 to solve for a²:
a² = 16
Now we're nearly there! If we find the square root of 16, we'll have our answer.
a = √16
Therefore, Side A = 4cm.
Practice geometry with DoodleMath
Want to learn more about the Pythagorean theorem
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Class 8 Courses
ABCD is a parallelogram in which P is the midpoint of DC and Q is a point on AC such that parallelogram in which $P$ is the midpoint of $D C$ and $Q$ is a point on $A C$ such that $C Q=\frac{1}{4} A C .$ If $P Q$ produced meets $B C$ at $R$, prove that $R$ is the midpoint $B C$.
Solution:
Join DB. We know that the diagonals of a parallelogram bisect each other. Therefore
$C S=\frac{1}{2} A C \quad \ldots$ (i)
Also, it is given that $C Q=\frac{1}{4} A C$ ...(ii)
Dividing equation (ii) by (i), we get:
$\frac{C Q}{C S}=\frac{\frac{1}{4} A C}{\frac{1}{2} A C}$
or, $C Q=\frac{1}{2} \operatorname{CS}$
Hence, Q is the midpoint of CS.
Therefore, according to midpoint theorem in $\triangle C S D$
$P Q \| D S$
if $P Q \| D S$, we can say that $Q R \| S B$
In $\triangle C S B, Q$ is midpoint of $C S$ and $Q R \| S B$.
Applying converse of midpoint theorem ,we conclude that R is the midpoint of CB. This completes the proof.
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Python Turtle Spiraling Shape
In this Python turtle tutorial, we will learn how to draw Python turtle spiraling shapes in Python and we will cover the different examples related to Turtle Spiraling shapes. And, we will cover these topics.
Python turtle spiraling star shape
In this part of the python turtle tutorial, we will learn how to draw a Python turtle spiraling star shape using turtle in python.
Star is a heavenly body which is visible at night it is also known as a luminous point it is very large but it seems like very small.
Here we can draw a spiral star with the help of turtle and turtle is work as a pen and we can draw the star on the screen and the screen is work as drawing board.
Code:
In the following code, we will import the turtle library from which we can draw a spiral star on the screen.
tur.title("Pythontpoint") is used to give the title to the window.
turt.forward(x * 20) is used to move the turtle in the forward direction.
Python turtle spiraling square shape
In this part of the python turtle tutorial, we will learn how to draw a Python turtlespiraling square shape in python.
Square has four equal sides and equal angles and make a plane figure and we draw a single square to moresquare as a spiral square.
Sprial square is draw with the the help of turtle as we know the turtle is work as a pen and this spiral square is drawn on the screen and screen is work as drawing sheet.
Code:
In this code, we will import the turtle library from which we can draw a spiral square with the help of a turtle.
tur.title("Pythontpoint") is used to give the title on the screen.
turt.forward(side) is used to move the turtle in the forward direction.
turt.right(90100):
turt.forward(side)
turt.right(90)
side = side - 2
Output:
After running the above code we get the following output in which we can see that the spiraling square is drawn on the screen.
draw spiraling square
Python turtle spiraling triangle shape
In this part of the python turtle tutorial, we will learn about how to draw a Python turtle spiral triangle shape in python.
Triangle has three sides and three angles and with these three sides a plane figure or spiraling triangle is drawn and here we draw a spiraling triangle with the help of a turtle.
Code:
In the following code, we will import turtle library from which we can draw a spiral triangle on the screen.
tur.title("Pythontpoint") is used to give the title to the screen.
turt.forward(side) is used to move the turtle in the forward direction.
turt.right(12070):
turt.forward(side)
turt.right(120)
side = side - 3
Output:
After running the above code we get the following code in which we can see that the spiraling triangle is drawn on the screen.
draw spiraling triangle
Python turtle spiraling pentagon shape
In this part of the python turtle tutorial, we will learn how to draw a python turtle pentagon shape in python.
Pentagon has five straight side and five angles from which a plane figure and a spiraling pentagon is drawn on the screen with the help of a turtle.
Code:
In the following code, we will import the turtle library from which we can draw a spiral pentagon on the screen.
turt.forward(side) is used to move the turtle in the forward direction.
turt.right(72) is used to move the turtle in the right direction.
from turtle import *
import turtle as tur
turt = tur.Turtle()
side = 200
for i in range(104):
turt.forward(side)
turt.right(72)
side = side - 2
Output:
After running the above code we get the following output in which we can see that the spiraling pentagon is drawn on the screen.
draw spiraling pentagon
Pythonturtle spiraling polygon shape
In this part of the python turtle tutorial, we will learn about how to draw a Python turtlespiraling polygon shape in python.
Polygon has at least three straight sides and three straight angles from which a plane figure or spiraling polygon is drawn on the screen.
Code:
In the following code, we will import the turtle library from which a spiral polygon is drawn on the screen.
turt.forward(length_Of_Side) is used to move the turtle in the forward direction.
turt.right(exteriorAngle) is used to move the turtle in the right direction.
from turtle import *
import turtle as tur
turt = tur.Turtle()
number_Of_Sides = int(input('Enter the number of sides of a polygon: '))
length_Of_Side = int(input('Enter the length of a side of a polygon: '))
exteriorAngle = 360/number_Of_Sides
for i in range(200):
turt.forward(length_Of_Side)
turt.right(exteriorAngle)
length_Of_Side = length_Of_Side - 0.5
Output:
After running the above code we get the following output in which we can see that the spiraling polygon is drawn on the screen.
draw spiraling polygon
So, in this tutorial, we discussed Python Turtle spiraling shapes and we have also covered different examples related to its implementation. Here is the list of examples that we have covered.
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NCERT Solutions for Class 10 Maths Constructions
We have provided NCERT Solutions for Class 10 Maths Ch 11 Constructions in this page which will be helpful in completing your homework in less time. It will encourage students to learn new topics and give better understanding of the chapter. Class 10 NCERT Solutions is the best way to revise whole chapter properly and learning efficiently. These solutions are updated according to the latest pattern of CBSE.
Page No: 219
Exercise 11.1
In each of the following, give the justification of the construction also: 1. Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts.
Answer
Steps of Construction: Step I: AB = 7.6 cm is drawn. Step II: A ray AX making an acute angle with AB is drawn. Step III: After that, a ray BY is drawn parallel to AX making equal acute angle as in the previous step. Step IV: Point A1, A2, A3, A4 and A5 is marked on AX and point B1, B2.... to B8 is marked on BY such that AA1 = A1A2 = A2A3 =....BB1= B1B2 = .... B7B8 Step V: A5 and B8 is joined and it intersected AB at point C diving it in the ratio 5:8.
AC : CB = 5 : 8
Justification:
ΔAA5C ~ ΔBB8C
∴ AA5/BB8 = AC/BC = 5/8
2. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle.
Answer
Steps of Construction: Step I: AB = 6 cm is drawn. Step II: With A as a centre and radius equal to 4 cm, an arc is draw. Step III: Again, with B as a centre and radius equal to 5 cm an arc is drawn on same side of AB intersecting previous arc at C. Step IV: AC and BC are joined to form ΔABC. Step V: A ray AX is drawn making an acute angle with AB below it. Step VI: 5 equal points (sum of the ratio = 2 + 3 =5) is marked on AX as A1 A2....A5 Step VII: A5B is joined. A2B' is drawn parallel to A5B and B'C' is drawn parallel to BC.
ΔAB'C' is the required triangle
3. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle.
Answer
Steps of Construction: Step I: A triangle ABC with sides 5 cm, 6 cm and 7 cm is drawn. Step II: A ray BX making an acute angle with BC is drawn opposite to vertex A. Step III: 7 points as B1 B2 B3 B4 B5 B6 and B7 are marked on BX. Step IV; Point B5 is joined with C to draw B5C. Step V: B7C' is drawn parallel to B5C and C'A' is parallel to CA.
Thus A'BC' is the required triangle.
4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1.5 times the corresponding sides of the isosceles triangle.
Answer
Steps of Construction: Step I: BC = 5 cm is drawn. Step II: Perpendicular bisector of BC is drawn and it intersect BC at O. Step III: At a distance of 4 cm, a point A is marked on perpendicular bisector of BC. Step IV: AB and AC is joined to form ΔABC. Step V: A ray BX is drawn making an acute angle with BC opposite to vertex A. Step VI: 3 points B1 B2 and B3 is marked BX. Step VII: B2 is joined with C to form B2C. Step VIII: B3C' is drawn parallel to B2C and C'A' is drawn parallel to CA.
Thus, A'BC' is the required triangle formed.
5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.
Answer
Steps of Construction:
Step I: BC = 6 cm is drawn. Step II: At point B, AB = 5 cm is drawn making an
∠ABC = 60° with BC. Step III: AC is joined to form ΔABC. Step IV: A ray BX is drawn making an acute angle with BC opposite to vertex A. Step V: 4 points B1 B2 B3 and B4 at equal distance is marked on BX. Step VII: B3 is joined with C' to form B3C'. Step VIII: C'A' is drawn parallel CA.
Thus, A'BC' is the required triangle.
7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.
Answer
Steps of Construction: Step I: BC = 3 cm is drawn. Step II: At B, A ray making an angle of 90° with BC is drawn. Step III: With B as centre and radius equal to 4 cm, an arc is made on previous ray intersecting it at point A. Step IV: AC is joined to form ΔABC. Step V: A ray BX is drawn making an acute angle with BC opposite to vertex A. Step VI: 5 points B1 B2 B3 B4 and B5 at equal distance is marked on BX. Step VII: B3C is joined B5C' is made parallel to B3C. Step VIII: A'C' is joined together.
Thus, ΔA'BC' is the required triangle.
Justification:
As in the previous question 6.
Page No. 221
Exercise 11.2
In each of the following, give also the justification of the construction: 1. Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
Answer
Steps of Construction: Step I: With O as a centre and radius equal to 6 cm, a circle is drawn. Step II: A point P at a distance of 10 cm from the centre O is taken. OP is joined. Step III: Perpendicular bisector OP is drawn and let it intersected at M. Step IV: With M as a centre and OM as a radius, a circle is drawn intersecting previous circle at Q and R. Step V: PQ and PR are joined.
Thus, PQ and PR are the tangents to the circle.
On measuring the length, tangents are equal to 8 cm.
PQ = PR = 8cm.
Justification:
OQ is joined.
∠PQO = 90° (Angle in the semi circle)2. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.
Answer
Steps of Construction:
Step I: With O as a centre and radius equal to 4 cm, a circle is drawn.
Step II: With O as a centre and radius equal to 6 cm, a concentric circle is drawn.
Step III: P be any point on the circle of radius 6 cm and OP is joined.
Step IV: Perpendicular bisector of OP is drawn which cuts it at M
Step V: With M as a centre and OM as a radius, a circle is drawn which intersect the the circle of radius 4 cm at Q and R
Step VI: PQ and PR are joined.
Thus, PQ and PR are the two tangents.
Measurement:
OQ = 4 cm (Radius of the circle)
PQ = 6 cm ( Radius of the circle)
∠PQO = 90° (Angle in the semi circle)
Applying Pythagoras theorem in ΔPQO,
PQ2 + QO2 = PO2
⇒ PQ2 + 42= 62
⇒ PQ2 + 16 = 36
⇒ PQ2 = 36 - 16
⇒ PQ2 = 20
⇒ PQ = 2√5
Justification:
∠PQO = 90° (Angle in the semi circle)
3. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.
Answer
Steps of Construction: Step I: With O as a centre and radius equal to 3 cm, a circle is drawn. Step II: The diameter of the circle is extended both sides and an arc is made to cut it at 7 cm. Step III: Perpendicular bisector of OP and OQ is drawn and x and y be its mid-point. Step IV: With O as a centre and Ox be its radius, a circle is drawn which intersected the previous circle at M and N. Step V: Step IV is repeated with O as centre and Oy as radius and it intersected the circle at R and T. Step VI: PM and PN are joined also QR and QT are joined.
Thus, PM and PN are tangents to the circle from P and QR and QT are tangents to the circle from point Q.
Justification:
∠PMO = 90° (Angle in the semi circle)
∴ OM ⊥ PM
Therefor, OM is the radius of the circle then PM has to be a tangent of the circle.
Similarly, PN, QR and QT are tangents of the circle.
4. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°.
Answer
We know that radius of the circle is perpendicular to the tangents.
Sum of all the 4 angles of quadrilateral = 360°
∴ Angle between the radius (∠O) = 360° - (90° + 90° + 60°) = 120°
Steps of Construction: Step I: A point Q is taken on the circumference of the circle and OQ is joined. OQ is radius of the circle. Step II: Draw another radius OR making an angle equal to 120° with the previous one. Step III: A point P is taken outside the circle. QP and PR are joined which is perpendicular OQ and OR.
Thus, QP and PR are the required tangents inclined to each other at an angle of 60°.
5. Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
Answer
Steps of Construction: Step I: A line segment AB of 8 cm is drawn. Step II: With A as centre and radius equal to 4 cm, a circle is drawn which cut the line at point O. Step III: With B as a centre and radius equal to 3 cm, a circle is drawn. Step IV: With O as a centre and OA as a radius, a circle is drawn which intersect the previous two circles at P, Q and R, S. Step V: AP, AQ, BR and BS are joined.
Thus, AP, AQ, BR and BS are the required tangents.
Justification:
∠BPA = 90° (Angle in the semi circle)
∴ AP ⊥ PB
Therefor, BP is the radius of the circle then AP has to be a tangent of the circle.
Similarly, AQ, BR and BS are tangents of the circle. 6. Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
Answer
Steps of Construction: Step I: A ΔABC is drawn. Step II: Perpendicular to AC is drawn to point B which intersected it at D.
Step III: With O as a centre and OC as a radius, a circle is drawn. The circle through B, C, D is drawn. Step IV: OA is joined and a circle is drawn with diameter OA which intersected the previous circle at B and E. Step V: AE is joined.
Thus, AB and AE are the required tangents to the circle from A.
Justification:
∠OEA = 90° (Angle in the semi circle)
∴ OE ⊥ AE
Therefor, OE is the radius of the circle then AE has to be a tangent of the circle.
NCERT Solutions for Class 10 Maths Chapter 11 Constructions
Class 11 Constructions NCERT Solutions will be very useful in preparation of the examinations and understanding the concepts properly. There are three exercises given in Chapter 11 Constructions which will embed with you knowledge about various types of mathematical constructions.
• Introduction: In earlier class, we have done certain constructions such as bisecting an angle, drawing the perpendicular bisector of a line segment, some constructions of triangles etc. and in this chapter, we will be doing more constructions like that and will give mathematical reasoning behind them.
• Division of a Line Segment: We will be dividing a line segment in a given ratio and construct a triangle similar to a given triangle as per given scale factor.
• Construction of Tangents to a Circle: We have to construct the tangents to a circle from a point outside it and provide justification of the construction also.
If you want to access other Study material for Class 10 then you can find them below.
There are only two exercises in the Chapter 11 Constructions which will improve your logical and thinking skills. We have also provided NCERT Solutions for Class 10 Maths exercisewise which you can get from below.
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Class ElevationAngle
The angular height of an object measured from the horizontal plane.
The elevation angle is part of local topocentric coordinates together with azimuth and distance.
For visible objects the elevation is an angle between 0° and 90°.
Note:Elevation angle and altitude angle may be used interchangeably.
Both altitude and elevation words are also used to describe the
height in meters above sea level.
ElevationAngle
Constructs a newly allocated ElevationAngle object that contain the angular value
represented by the string. The string should represent an angle in either fractional degrees
(e.g. 45.5°) or degrees with minutes and seconds (e.g. 45°30').
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The diagonals of the rectangle intersect at an angle of 60 degrees. The length of the diagonal is 12.
The diagonals of the rectangle intersect at an angle of 60 degrees. The length of the diagonal is 12. Find the length of the larger side of the rectangle.
Crossing, the diagonals form a triangle with the smaller side of the rectangle. This is an isosceles triangle, because the diagonals are halved in the rectangle.
Let us determine what the angles at the base (smaller side) will be equal to when the angle between the diagonals is 60 °:
(180 – 60): 2 = 60.
That is, all three angles are 60 °. Then the smaller side is equal to half the diagonal:
12: 2 = 6.
Let's find out what the area of the rectangle will be:
1/2 * 12 ^ 2 * √3 / 2 = 36√3.
Let's find out the length of the longer side:
36√3: 6 = 6√3.
Answer: 6√
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In an acute-angled triangle $ABC$, a point $D$ lies on the segment $BC$. Let $O_1$,$O_2$ denote the circumcentres of triangles $ABD$ and $ACD$, respectively. Prove that the line joining the circumcentre of triangle $ABC$ and the orthocentre of triangle $O_1O_2D$ is parallel to $BC$.
Supposing that the circumcentre of $\Delta$$ABC$ is $O$, and the orthocenter of $\Delta$$O_1O_2D$ is H, I could prove that $A,O_1,O,H,O_2$ lie on a circle. After that I cannot figure out how to do. Please help.
$\begingroup$This is not a site like Brilliant or AOPS. Here, the questioner must show effort and then the answer is to help him/her to complete the solution. One has to draw a diagram to solve the problem. Hence, the question is easier to comprehend with a diagram. As for the fact that multiple configurations are possible, draw any possible configuration.$\endgroup$
$\begingroup$I do understand, @Anubhab Ghosal. I am not simply posting questions or asking doubts. I do write whatever I have done as a part of my attempt to solve the problem. But please see, I registered with this website just yesterday and I am yet to explore and learn more on this website. So maybe in my future posts I will try my best to post a figure, along with the question, too. Anyways, thanks for the suggestions!$\endgroup$
1 Answer
1
I proceded starting from your suggestions and then by the following path.
You can use your result to state that
\begin{equation}
\angle O_2OH \cong\angle O_2AH \cong \angle O_2DH,\tag{1}\label{eq:cong1}
\end{equation}
where the first congruence is due to the fact that the angles are subtended by the same chord $O_2H$, and the second congruence is due to the fact that triangle $AO_2D$ is isosceles. Call then $M_1$ the middle point of $BC$ and $M_2$ the middle point of $AC$. Note that
\begin{equation}
\angle O_2OM_1 + \angle ACB \cong \pi,\tag{2}\label{eq:sum1}
\end{equation}
which can be obtained by adding up the internal angles of quadrilateral $CM_2OM_1$. Now subtract and add $\angle O_2OH$ on the LHS of \eqref{eq:sum1}, getting
\begin{eqnarray}
\angle O_2OM_1 -\angle O_2OH+ \angle ACB + \angle O_2OH&\cong& \pi\\
\angle HOM_1 + \angle ACB + \angle O_2OH \cong \pi.
\end{eqnarray}
Finally, by using \eqref{eq:cong1} and properties of the circumcenter $O_2$, show that
\begin{equation}
\angle ACB + \angle O_2OH \cong \frac{\pi}{2},
\end{equation}
which will lead to the thesis.
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Figure Of Pentagon
Figure Of Pentagon - Sum of exterior angles is 360°. This shape must have five sides that form a closed 2d figure. A pentagon may be either. (a closed figure is one in which all of its sides meet with each other to. Septagon (7 sides) think septagon is. Web the regular pentagram has a special number hidden inside called the golden ratio, which equals approximately 1.618 a/b = 1.618. Web updated october 19, 2023 at 8:19 p.m. Depending on the characteristics that we. Web here, n represents the total number of sides in a pentagon. Edt | published october 19, 2023 at 7:47 a.m.
So You Want to Know About Pentagons? Math ∞ Blog
In a pentagon, an angle is formed. Web the pentagon is a geometric shape that is characterized by its five sides and five angles. Hexagon (6 sides) honeycomb has hexagons. Web the sum of all angles is determined by the following formula for a polygon: It is a polygon, which.
Geometric Shape, Pentagon Shape, shapes, Pentagon, Geometric Figure
It is an equilateral polygon, meaning that all of its sides are. Web updated october 19, 2023 at 8:19 p.m. Web a pentagon that has equal sides and equal interior angles is known as a regular pentagon. Web oct 20, 2023 at 11:44 am edt. A regular pentagon has 5 equal edges and 5 equal angles.
Web the pentagon is a geometric shape that is characterized by its five sides and five angles. Septagon (7 sides) think septagon is. Sum of exterior angles is 360°. Web in the aftermath of the deadly hospital explosion in gaza on tuesday, some of the world's biggest and most reputable. A pentagon may be either.
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Web the pentagon is a geometric shape that is characterized by its five sides and five angles. The path forward for house. It is a polygon, which. Septagon (7 sides) think septagon is. Web the sum of all angles is determined by the following formula for a polygon:
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Web the sum of all angles is determined by the following formula for a polygon: In a pentagon, there are 5 sides, or. A regular pentagon has 5 equal edges and 5 equal angles. Web properties of a pentagon. Sum of interior angles is 540°.
Set colorful stitched pentagon shape Royalty Free Vector
Web the pentagon is a geometric shape that is characterized by its five sides and five angles. Septagon (7 sides) think septagon is. Web pentagon (5 sides) the pentagon in washington dc has 5 sides. A regular pentagon has 5 equal edges and 5 equal angles. Observe the following figure which shows the exterior angles of a.
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Web a blast at a gaza hospital killed hundreds of palestinians just before u.s. Web pentagon (5 sides) the pentagon in washington dc has 5 sides. It is an equilateral polygon, meaning that all of its sides are. 19, kick signed massively popular. Sum of interior angles is 540°.
Pentagon Definition, Shape, Properties, Formulas
Web in the aftermath of the deadly hospital explosion in gaza on tuesday, some of the world's biggest and most reputable. Web the sum of all angles is determined by the following formula for a polygon: Web here, n represents the total number of sides in a pentagon. Web a pentagon that has equal sides and equal interior angles is.
Secrets of a Certain Pentagon
Web here, n represents the total number of sides in a pentagon. Observe the following figure which shows the exterior angles of a. Web the sum of all angles is determined by the following formula for a polygon: Learn more about pentagon, regular and. This shape must have five sides that form a closed 2d figure.
Seriously! 38+ Little Known Truths on Pentagon Shape 4 Sides You can
Web a blast at a gaza hospital killed hundreds of palestinians just before u.s. It is an equilateral polygon, meaning that all of its sides are. Web updated october 19, 2023 at 8:19 p.m. Septagon (7 sides) think septagon is. It is a polygon, which.
Sum of exterior angles is 360°. Web in the aftermath of the deadly hospital explosion in gaza on tuesday, some of the world's biggest and most reputable. President joe biden visited israel as. Web a blast at a gaza hospital killed hundreds of palestinians just before u.s. Edt | published october 19, 2023 at 7:47 a.m. (a closed figure is one in which all of its sides meet with each other to. Depending on the characteristics that we. In a pentagon, there are 5 sides, or. Sum of interior angles is 540°. Observe the following figure which shows the exterior angles of a. In a pentagon, an angle is formed. Septagon (7 sides) think septagon is. Learn more about pentagon, regular and. Web a pentagon that has equal sides and equal interior angles is known as a regular pentagon. Web pentagon (5 sides) the pentagon in washington dc has 5 sides. A regular pentagon has 5 equal edges and 5 equal angles. It is a polygon, which. Web updated october 19, 2023 at 8:19 p.m. Web the regular pentagram has a special number hidden inside called the golden ratio, which equals approximately 1.618 a/b = 1.618. Web properties of a pentagon.
Web The Sum Of All Angles Is Determined By The Following Formula For A Polygon:
It is an equilateral polygon, meaning that all of its sides are. It is a polygon, which. Depending on the characteristics that we. Web here, n represents the total number of sides in a pentagon.
Web The Pentagon Is A Geometric Figure That Has Five Sides And Five Angles.
This shape must have five sides that form a closed 2d figure. Web in the aftermath of the deadly hospital explosion in gaza on tuesday, some of the world's biggest and most reputable. Web the pentagon is a geometric shape that is characterized by its five sides and five angles. Web a pentagon that has equal sides and equal interior angles is known as a regular pentagon.
Web Oct 20, 2023 At 11:44 Am Edt.
Web updated october 19, 2023 at 8:19 p.m. Edt | published october 19, 2023 at 7:47 a.m. Hexagon (6 sides) honeycomb has hexagons. The interior angles in a pentagon add up to 540°.
Sum Of Exterior Angles Is 360°.
President joe biden visited israel as. Web the regular pentagram has a special number hidden inside called the golden ratio, which equals approximately 1.618 a/b = 1.618. In a pentagon, there are 5 sides, or. The path forward for house.
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Find the area of the parallelogram with vertices A(-3, 0), B(-1, 5), C(7, 4), and D(5, -1)
The aim of this problem is to get us familiar with the area of a very common quadrilateral known as a parallelogram. If we recall, a parallelogram is a pretty simple quadrilateral with two couples of parallel-faced sides.
The opposite lengths of a parallelogram are of equal dimensions and the opposing angles of a parallelogram are of equal magnitude.
Expert Answer
Since a parallelogram is a tilted rectangle, all of the area formulas for known quadrilaterals can be used for parallelograms.
A parallelogram with one base $b$ and height $h$ can be separated into a trapezoid and a triangle with a right-angled side and can be shuffled into a rectangle. This implies that the area of a parallelogram is identical to that of a rectangle which has the same base and height.
We can define the area of a parallelogram as the absolute magnitude of the cross product of its adjacent angles, that is:
\[Area = |\overline{AB} \times \overline{AD}|\]
Finding the adjacent edges $\overline{AB}$ and $\overline{AD}$ and substituting back in the equation as follows:
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NCERT Solutions for CBSE Class 9 Maths Chapter 7
NCERT Solutions for Class 9 Maths Chapter 7Triangles provides the solutions and problems related to the chapter as included in the CBSE Syllabus.
Class 9 Maths NCERT Solutions Chapter 7 Triangles in PDF format available as below. Students can download the free PDF of these NCERT Solutions for Class 9 by clicking on the link below to keep it handy for future reference.
What is the meaning of congruence of triangles in NCERT Solutions for Class 9 Maths Chapter 7?
According to NCERT Solutions for Class 9 Maths Chapter 7, congruence of triangles states that two triangles are said to be congruent if they are copies of each other, and when superposed, they cover each other exactly. In other words, two triangles are congruent if the sides and angles of one triangle are equal to the corresponding sides and angles of the other triangle.
Practising NCERT Solutions for Class 9 Maths Chapter 7 helps the students to top the CBSE exams and ace the subject. These solutions are devised, based on the most updated CBSE syllabus, covering all the crucial topics of the respective subjects. Hence, solving these questions will make the students more confident to face the board exams. Topics given in these solutions form the basis for top scores. It also helps the students to get familiar with answering questions of all difficulty levels. These solutions are highly recommended to the students for reference and to practise for the CBSE exams.
We hope this information on "NCERT Solution Class 9 Maths Chapter 7" is useful for students. Click above link of NCERT Solutions to get the solved answers from the NCERT book for all the problems of chapter 7. Stay tuned for further updates on CBSE and other competitive exams. To access interactive Maths and Science Videos, download ANAND CLASSES's App.
Triangles is a very important chapter as the fundamental concepts introduced here are carried forward to higher levels of education. For this purpose, after completing the questions from the NCERT Textbook, students are recommended to solve the other textbooks which are prescribed by the CBSE Board.
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Chukwuemeka
Past Exam Questions, Solutions, and Explanations on Circle Theorems JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.
For WASSCE Students
Questions with multiple choice options are for the WASSCE General Mathematics Paper 1 (Objective)
Questions without multiple choice options are for the WASSCE General Mathematics Paper 2 (Theory)
For GCSESolve all questions.
State the reason(s) for each step.
Show all work.
Some of these questions can be solved in at least two (two or more) ways.
Teachers can encourage students to get the same solution using more than one approach/method.
Where applicable, depending on time; I solved some questions using more than one approach.
(1.) CSEC The diagram below shows a circle where $AC$ is a diameter.
$B$ and $D$ are two other points on the circle and $DCE$ is a straight line.
Angle $CAB = 28^\circ$ and $\angle DBC = 46^\circ$
Calculate the value of each of the following angles.
Show detailed working where necessary and give a reason to support your answers.
(6.) GCSE This circle has centre C. W, X and Y are points on the circle. WY is a straight line.
Tick one box for each statement.
WY is a diameter ... True
This is because it is a chord that passes through the centre of the circle.
WX is a radius. ... False
It does not pass through the centre of the circle.
The shaded section is a sector. ... False
It is a segment. It is the minor segment.
A sector will also include the area enclosed by both radii: CW and CX
In other words, the minor sector will be the area of ▵WCX and the area of minor segment WX
Arc XY is part of the circumference. ... True
Arcs of a circle are part of the circumference of the circle.
$
(7.) CSEC The diagram below, not drawn to scale, shows a circle with centre $O$.
$HJ$ and $HG$ are tangents to the circle and $\angle JHG = 48^\circ$
Calculate, giving the reason for each step of your answer, the measure of:
(10.) GCSE The diagram shows a circle with centre O. P, Q and R are points on the circle.
Choose words from the box to complete these sentences.
(i) Line OP is a ........................
(ii) Line QR is a ........................
(iii) Lines OP and QR are ........................
(i) Line OP is a radius
(ii) Line QR is a chord
(iii) Lines OP and QR are parallel because of the similar single arrows
(11.) JAMB $PQRS$ is a cyclic quadrilateral in which $PQ = PS$.
$PT$ is a tangent to the circle and $PQ$ makes an angle of $50^\circ$ with the tangent as shown in the
figure below.
What is the size of $QRS$?
(13.) CSEC The diagram below, not drawn to scale, shows a circle with centre $O$.
The vertices $H, J, K\:\:and\:\:L$ of a quadrilateral lie on the circumference of the circle and $PKM$ is a
tangent to the circle at $K$.
The measure of angle $H\hat{J}L = 20^\circ$ and $J\hat{K}H = 50^\circ$
Calculate, giving reasons for each step of your answer, the measure of
(17.) CSEC The diagram below, not drawn to scale, shows a circle with centre $O$.
The points $A, B, C\:\:and\:\:D$ are on the circumference of the circle.
$EAF$ and $EDG$ are tangents to the circle at $A$ and $D$ respectively.
$A\hat{O}D = 114^\circ$ and $C\hat{D}G = 18^\circ$
Calculate, giving reasons for EACH step of your answer, the measure of
(21.) CSEC (a) In the diagram below, $VWZ$ and $WXYZ$ are two circles intersecting at $W$ and $Z$.
$SVT$ is a tangent to the circle at $V$
$VWX$ and $VZY$ are straight lines.
$T\hat{V}Y = 78^\circ$ and $S\hat{V}X = 51^\circ$
(i) Calculate the size of EACH of the following angles, giving reasons for your answers.
(22.) ACT The circle below has its center at O, and points A and B lie on the circle.
The length of $\overline{AO}$ is 4 inches, and arc $\overset{\huge\frown}{AB}$ measures 112°
What is the measure of ∠ABO?
In the diagram above, not drawn to scale, $O$ is the centre of the circle.
$\angle AOB = 130^\circ$, $\angle DAE = 30^\circ$, and $AEC$ and $BED$ are chords of the circle.
(a) Calculate the size of EACH of the following angles, giving reasons for EACH step of your answers.
(32.) GCSEP, Q, R and S are points on a circle. PXR and QXS are straight lines. PX = SX
Prove that QS is not a diameter of the circle.
If QS is the diameter of the circle, then $\angle QPS = 90^\circ...\angle\;\;in\;\;a\;\;semicircle$
If QS is not the diameter of the circle, then $\angle QPS \ne 90^\circ$
So, let us calculate $\angle QPS$
A, B and C are points on the circumference of a circle, centre O DAE is the tangent to the circle at A
Angle BAE = $56^\circ$
Angle CBO = $35^\circ$
Work out the size of angle CAO
You must show all your working.
TV is a tangent to the circle at P
SR = RQ
Angle QPV = 41° and angle SQP = 57°
Show that SP is parallel to RQ
You must give reasons to justify any angles that you calculate.
We can review the properties of angles between parallel lines to determine that SP is parallel to RQ
Of those properties, the easiest one to check for is the Alternate Interior Angles property...alternate interior angles
between parallel lines are equal
Let us demonstrate it with a diagram
So, we need to show that $\angle RQS = \angle PSQ$ ...indicated by the angles marked in red
Once, we can show that $\angle RQS = \angle PSQ$, then we have shown one of the properties of angles between parallel lines
and hence show that the two lines: SP and RQ are parallel
(47.) NSC In the diagram, KLMN is a cyclic quadrilateral with $K\hat{L}M = 87^\circ$
Diagonals LN and MK are drawn.
P is a point on the circle and MP is produced to T, a point outside the circle.
Chord LP is drawn.
$L\hat{M}K = y$ and $\hat{N_1} = 2y$
A, B, C and D are points on a circle.
The line PQ is a tangent to the circle at B
$P\hat{B}A = 51^\circ$, $B\hat{C}D = 118^\circ$ and $A\hat{B}D = w^\circ$
Find the value of w
You must show all your working.
(b.)
E, F and G are points on a circle with centre H
$E\hat{F}H = x$ and $G\hat{F}H = y$
Complete the proof of the following statement:
Tha angle at the centre is twice the angle at the circumference.
Proof:
$E\hat{F}G = x + y$
$F\hat{E}H = x$ (triangle FEH is isosceles)
$F\hat{H}E = .................................$
(54.) ACT A circle with center C is shown below.
Points W, X, Y, and Z lie on the circle.
The measure of $\angle WCY$ is $100^\circ$, the measure of $\angle XCZ$ is $80^\circ$, and the measure of $\angle WCZ$ is
$150^\circ$
What is the measure of $\angle XCY$?
(60.) NZQA In this spider web, points G, H, T, and S all lie on the circumference of a Circle, with centre C
The straight lines FSJ and FTK are both tangents to the circle at the points S and T
Angle SCT = 100°
Determine whether the line FSJ is parallel to the line GCT Justify your answer with clear geometrical reasoning
$
\underline{\triangle COD} \\[3ex]
\angle COD = 90^\circ ...implies\;\;a\;\;right\;\;\triangle \\[3ex]
$
A right triangle is a triangle that has one angle of $90^\circ$ and two angles less than $90^\circ$
Option D appears in the figure
The only remaining option is Option E
(69.) GCSEA, B and C are points on a circle. CD is a tangent to the circle.
We observe at least two circle theorems with this circle:
(1.) Angle in a semicircle: we will need to do some construction to join point B to point E before we can use this theorem
(2.) The figure inside the circle is a cyclic quadrilateral. Hence, we shall use at least one of the theorems dealing with
cyclic quadrilaterals
(72.) CSEC (a.) The circle shown below has centre O and the points A, B, C and D lying on the
circumference.
A straight line passes through the points A and B
Angle CBD = 49° and angle OAB = 37°
(i) Write down the mathematical names of the straight lines BC and OA BC ....................................................... OA .......................................................
(ii) Determine the value of EACH of the following angles.
Show detailed working where necessary and give a reason to support your answer.
(a) x
(b) y
(73.) USSCE Advance Mathematics Paper 2
(a.) Fill in the missing words.
The angle between a tangent and a chord through the point of contact is equal to the angle $\rule{4cm}{0.15mm}$ by the chord in the
alternate segment.
(b.) Two circles intersect at points P and Q
Line APB is drawn through P
The tangents at A and B meet at C
If $\angle ABC = \beta^\circ$ and $\angle BAC = \alpha^\circ$, find an expression in terms of $\alpha$ and $\beta$ for
(74.) ACT In the figure below, lines l and m are tangent to the circle at points B and D,
respectively.
Points A and C are on the circle.
The measure of $\angle ABC$ is $95^\circ$ and the measure of $\angle BCD$ is $85^\circ$
The lines in which of the following pairs of lines are necessarily parallel?
(79.) CSEC The diagram below, not drawn to scale, shows a circle.
The points P, Q, R, T and V are on the circumference. QRS is a straight line.
Angle PVR = 75° and angle TRS = 60°
Determine the value of EACH of the following angles.
Show detailed working where necessary and give a reason to support your answer.
(i) Angle PTR
(ii) Angle TPQ
(iii) Obtuse angle POR where O is the centre of the circle.
$
\angle TPQ = 60^\circ ...exterior\;\;\angle\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;PTRQ = interior\;\;opposite\;\;\angle \\[3ex]
$
Construction:
(a.) Indicate the centre of the circle, O
(b.) Draw the line from point P to centre O
(c.) Draw the line from centre O to point R
(80.) ACT In the figure below, line q is parallel to line r, C is the center of the circle.
$\overline{AE}$ and $\overline{BD}$ both go through C, q is tangent to the circle at B, F lies on
$\overleftrightarrow{AD}$, and the measure of $\angle FAC$ is $145^\circ$
What is the measure of $\angle BCE$?
(88.) ACT In the circle shown below, chords $\overline{TR}$ and $\overline{QS}$ intersect at P, which is the
center of the circle, and the measure of $\angle PST$ is $30^\circ$
What is the degree measure of minor arc $\overset{\huge\frown}{RS}$?
(96.) ACT The circle shown below has diameter $\overline{AD}$, and points B and C lie on the circle.
The measure of $\angle CAD$ is $30^\circ$, and the measure of minor arc $\overset{\huge\frown}{CD}$ is $60^\circ$
What is the measure of minor arc $\overset{\huge\frown}{AC}$
(112.) ACT From point A outside a circle and in the same plane as the circle, 2 rays are drawn tangent to the
circle with the points of tangency labeled B and C, respectively.
Segment $\overline{BC}$ is then drawn to form $\triangle ABC$
If $\angle A$ measures $70^\circ$, what is the measure of $\angle ABC$?
In the diagram, $\overline{TS}$ is a tangent to the circle at S
If O is the centre of the circle, $\angle TSP = 21^\circ$ and $\angle RQP = 100^\circ$, find with reasons:
(i) $\angle SPR$
(ii) $\angle QSR$
The diagram shows a circle, centre O, with points B and D on the circumference
The line AC touches the circle at B OB is parallel to DC and angle OAB = 49° (i) Write down the mathematical name of the line OB (ii) Write down the reason why angle ABO is 90° (iii) Find angle AOB (iv) Write down the reason why angle ADC = angle AOB (v) Complete the statement using a mathematical word.
Triangle AOB is ..................................... to triangle ADC (vi)AB = 5.4 cm
Calculate (a)OB (b)OA (c) the area of triangle AOB
(120.) NZQA Amy draws another pentagon which is cyclic but not regular.
PT is parallel to EN
PXT is a diameter
X is the centre of the circle
(i) Amy thinks that angle PAT is 90°
Is she correct?
Give a geometric reason.
(ii) Calculate the size of angle NXE
Give geometric reasons.
(iii) Suppose angle XPE = w
It can be shown that angle NXE = 4w = 180°
From this formula, what does this tell you about the size of angle w?
$
(i) \\[3ex]
Yes,\;\;she\;\;is\;\;correct \\[3ex]
\angle PAT = 90^\circ...\angle\;\;in\;\;a\;\;semicircle\;\;is\;\;a\;\;right\;\;\angle \\[3ex]
$
Construction: Draw the radii: from point X on the centre to point E on the circumference; and from point X on the centre to
point N on the circumference of the circle
(137.) CSEC The diagram below, not drawn to scale, shows a circle, with centre O
The points A, B, C and M are on the circumference.
The straight line CN is a tangent to the circle at the point C and is perpendicular to BN
The diagram is a circle passing through the points A, B, C and D such that AC and BD meet at a point
E inside the circle.
If ∠DAC = 27°, ∠ABD = 54° and ∠ACB = 63°, find:
(i) ∠CAB
(ii) ∠AEB
In the diagram, the two circles intersect at X and Y
The centre of the smaller circle is on the circumference of the bigger circle. A and B are any two points on the major arcs, one on each circle.
Find an equation connecting a and b
Construction: Draw two radii: one radius from point X to point O and the other radius from point Y to
point O
The above diagram shows a circle centre O where A, B, C and D are four points on the circle.
TD is a tangent touching the circle at D.
Calculate the angles a, b, c, d and e.
Give reasons and show your working where necessary.
(175.) NZQA Nails Q, V, and W all lie on the circumference of a circle, with centre C
Other nails form a straight line PQR which is tangent to the circle at point Q
The lines PVW and QCW are straight.
Angle QPV = 40°
AC is a tangent to the circle, centre O, with point of contact B.
DE is a diameter of the circle and F is a point on the circumference.
Angle ABD is 77° and angle DEF is 64°
Calculate the size of angle BDF
We can solve this question in at least two ways.
Use any approach you prefer.
(200.) ACT Right triangle $\triangle ABC$ is inscribed in a circle with center M, as shown below, and C can
be any point on the circle other than A or B
Which of the following is the most direct explanation of why $\triangle MCA$ is isosceles?
(209.) CSEC The diagram below shows a circle, with the points P, Q, R and S lying on its
circumference and its centre marked O RP is a diameter of the circle and AB is a tangent to the circle at P
Angle APQ = 3x°, angle QPR = 2x°, angle RPS = x° and angle QSP = 54°
Determine the value of EACH of the following angles.
Show detailed working where possible and give a reason for your answer.
(i.) x
(ii.) y
(iii.) z
(215.) NYSED Regents Examination In circle O two secants, $\overline{ABP}$ and $\overline{CDP}$, are
drawn to external point P.
If m$\overset{\huge\frown}{AC} = 72^\circ$, and m$\overset{\huge\frown}{BD} = 34^\circ$, what is the measure of
∠P?
B, C, D and E are points on a circle. AB is the tangent at B to the circle. AB is parallel to ED
Angle ABE = 73°
Work out the size of angle DCE
Give a reason for each stage of your working.
We have a chord and a tangent, so we may need the Alternate Segment Theorem
Let us do some construction Construction:
(1.) Draw a chord from Point B to Point D ...so we can use the Alternate Segment Theorem
(2.) Draw a chord from Point C to Point D ...we may need to use another Theorem
A and B are points on the circumference of a circle, centre O.
AC is a tangent to the circle.
Angle BAC = 2x
Find the size of the angle AOB, in terms of x, giving a reason for each stage of your working.
We can solve this question using at least two approaches
Use any approach you prefer.
Figure 1 shows the circle ABCD with centre O and diameter DC
The point T is such that TCOD is a straight line and TA is the tangent to the circle at A AT = 10cm TC = 8cm
(a) Calculate the radius, in cm, of the circle.
(b) Calculate the length, in cm to 3 significant figures, of the arc ABC
A, B, C and D are points on the circumference of the circle.
The line XY is a tangent to the circle at A.
(a) Find the value of x, giving a reason for your answer.
(b) Find the value of y, giving a reason for your answer.
(257.) ICSE In the given figure PQ is a tangent to the circle at A, AB and AD are bisectors of ∠CAQ and
∠PAC
If ∠BAQ = 30°, prove that:
(i) BD is a diameter of the circle
(ii) ABC is an isosceles triangle
(260.) ICSE In the given figure, ABCDE is a pentagon inscribed in a circle such that AC is a diameter and side
BC // AE.
If ∠BAC = 50°, find giving reasons:
(i) ∠ACB
(ii) ∠EDC
(iii) ∠BEC
Hence prove that BE is also a diameter
(261.) KCSE In the figure below, points A, B, C, D and E lie on the circumference of a circle centre O
Line FAG is a tangent to the circle at A
Chord DE of the circle is produced to intersect with the tangent at F
Angle FAE = 30°, ∠EDC = 110° and ∠OCB = 55°
(281.) KCSE In the figure below A, B, C and D are points on the circumference of the circle, centre O.
A tangent to the circle at A intersects chord CD produced at E.
Line AB is parallel to line EC.
Angle AED = 45° and angle ABD = 40°.
(a) Calculate the size of:
(i) ∠ADB
(ii) ∠OCD
(b) Given that ED = 3.5 cm and DC = 4.9 cm, calculate correct to 1 decimal place:
(i) the length of the tangent AE
(ii) the radius of the circle.
(288.) ACT Radius $\overline{OA}$ of the circle shown below is perpendicular to $\overline{AP}$
The circle intersects $\overline{OP}$ at B.
The length of $\overline{AP}$ is 12 centimeters, and the measure of ∠APO is 20°.
Which of the following values is closest to the length, in centimeters, of $\overline{BP}$ ?
(297.1) The exterior angle of a cyclic quadrilateral is equal to the interior opppsite angle.
$
(297.2.1) \\[3ex]
M\hat{P}R = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle\;\;is\;\;a\;\;right\;\;\angle \\[3ex]
(297.2.2) \\[3ex]
\hat{P_3} = 11^\circ ...given \\[3ex]
\hat{M_2} = \hat{P_3} = 11^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex]
\hat{L_4} = \hat{M_2} = 11^\circ ...alternate\;\;\angle s\;\;between\;\;parallel\;\;lines:|LN| \;\;and\;\; |RM|\;\;are\;\;equal \\[3ex]
\hat{P_1} = \hat{L_4} = 11^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex]
$
(297.2.3)
LN || RM ...given
Because parallel lines are equidistant from each other:
LR = NM because each point on the line LN is the same distance as the corresponding point on the other line RM
Hence the point L on the line LN is equidistant from the point R on the line RM
Also, the point N on the line LN is equidistant from the point M on the line RM
(299.) KCSE In the figure below, P, Q, R and S are points on the circle with centre O.
PRT and USTV are straight lines.
Line USTV is a tangent to the circle at S.
∠RST = 50° and ∠RTV = 150°
(301.) NSC In the diagram, TSR is a diameter of the larger circle having centre S.
Chord TQ of the larger circle cuts the smaller circle at M.
PQ is a common tangent to the two circles at Q.
SQ is drawn.
RP ⊥ PQ and MS || QR.
Let us represent the information in the circle Construction:
(a.) Let O be the center of the circle.
(b.) Indicate the central angles corresponding to the minor arcs.
(c.) Indicate the congruent lines (chords)
(309.) curriculum.gov.mt (a.) WXYZ is a cyclic quadrilateral.
XY and WZ are parallel.
∠WXY = 84°.
Work out the size of ∠XYZ.
(b.) Points B, C, D and E lie on the circumference of circle centre O and line AE is a tangent to the circle at E.
∠BDE = 35°
Work out the value of the angles marked x, y and z, giving reasons for your
answers.
(a.)
For us to use the one of the properties of angles between parallel lines: Construction:
Extend Line YX to A
Extend Line ZW to B
(316.) GCSE The diagram shows a circle with centre O.
Points A, B, C and D all lie on the circumference of th circle.
The radius of the circle is 3.6cm, BC = 4.1 cm and $B\hat{C}D$ = 93°.
Prove that $D\hat{B}C$ = 52.3°, correct to 3 significant figures.
You must show all your working and give a reason for each stage of your proof.
(320.) ACT Graphed in the standard (x, y) coordinate plane below is line l and the circle with
equation $(x - 2)^2 + y^2 = 1$.
Line l passes through O(0, 0) and is tangent to the circle at A, and B is the center of the
circle.
What is the measure of $\angle AOB$?
(323.) NZQA The points P, Q, R, and S all lie on the circumference of circle, with centre C.
TP and TQ are tangents to the circle.
PCR is a diameter of the circle.
Angle PSQ is 68°. RS = 10 cm. PS = 16 cm.
A, C and D are points on a circle, centre O. AB and CB are tangents to the circle.
Angle ABC = 74°
Work out the size of angle ADC
Show your working clearly.
Construction: To use these theorems: Intersecting Tangents Theorem and Angle of Intersecting Tangents Theorem,
we will need to draw the line from the centre of the circle to the intersection of the two tangents: Point B
A and B are points on the circumference of a circle, centre O CBD is the tangent to the circle at B CAO is a straight line.
∠ABC = 38°
Giving your reasons, find the size, in degrees, of ∠ACB
Figure 6 shows a triangle ABC and a circle PQR, centre O.
The triangle is such that side AB is the tangent to the circle at P, side BC is the tangent to the
circle at Q and side AC is the tangent to the circle at R.
The region inside the circle is shaded, as shown in Figure 6. AB = 14cm, BC = 12cm and AC = 8cm
Let BP = xcm and by considering the lengths of the tangents to the circle,
(a) obtain an equation in x only and solve it to find the length, in cm, of BP.
(b) Find, to 3 significant figures, the area of the circle as a percentage of the total area of triangle ABC.
(a) Construction: To get a clearer picture of x and the remaining lengths, let us indicate the
length of x and the radii.
Let:
|AP| = m
|CQ| = n
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Angle Pairs Definition
Definition of angle pairs
The vertical angles are defined as two non-adjacent angles formed by intersecting lines. Angles of vertical opposition - Conceptual understanding of the subject:
Sequential set for inner angle
Cutting two straight line through a transverse is done by calling the angle pairs on one side of the transverse and within the two straight line the successive inner angle. Shown in the illustration are the angle 3 and 5 successive inner angle. Also, the angle 4 and 6 are successive inner angle.
In the case where two parallels are intersected by a transverse, the pairs of successive inner corners are complementary. Evidence: 3 and 5 are complementary and 4 and 6 are complementary. k 8 is a traversing.
Allow your pupils to learn some mathematics and show their artistry at the same time! Most of my pupils like to show their other talent while fulfilling their mathematical duties. So will your disciples! There are 12 issues with specific angle pairs. Issues demand understanding of the concept of perpendicular angle, additional angle, complement angle, neighboring angle, and pairing.
The student will find the specified value for each query. Use TPT balance for your next purchases: Pleas go to your My Shopping page. Next to each sale you will see a Give me your opinion panel. Every and every times you give your input, TPT gives you feed-back vouchers that you use to reduce the costs of your upcoming purchasing.
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Understanding the SAS Criterion: Proving Triangle Congruence with Side, Angle, and Side
SAS
SAS is an abbreviation used in geometry to describe a specific method for proving that two triangles are congruent
SAS is an abbreviation used in geometry to describe a specific method for proving that two triangles are congruent. The letters S, A, and S stand for Side, Angle, and Side, respectively.
To prove that two triangles are congruent using the SAS criterion, three conditions must be met:
1. The lengths of two sides of one triangle must be equal to the lengths of two corresponding sides of the other triangle.
2. The measures of the included angles (the angle between the two sides) must be equal.
3. The remaining side and angle of one triangle must have the same lengths and measures as the remaining side and angle of the other triangle.
To provide a more detailed explanation of the SAS criterion, consider the following hypothetical scenario:
Let's say we have two triangles, triangle ABC and triangle DEF. To prove that these triangles are congruent using SAS, we need to demonstrate that two pairs of sides and the included angle are equal.
1. Side AB = side DE: The length of side AB in triangle ABC is equal to the length of side DE in triangle DEF.
2. Side BC = side EF: The length of side BC in triangle ABC is equal to the length of side EF in triangle DEF.
3. Angle B = angle E: The included angle, angle B in triangle ABC, is equal to the included angle, angle E in triangle DEF.
If these conditions are satisfied, we can conclude that triangle ABC is congruent to triangle DEF, using the SAS criterion.
Remember, the SAS criterion is just one of several methods for proving congruence between triangles. Other methods include Side-Side-Side (SSS), Angle-Angle-Side (AAS), and Hypotenuse-Leg (HL) for right triangles. It is essential to use the appropriate criterion
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How do you recognize a central angle of a circle
Find an answer to your question 👍 "How do you recognize a central angle of a circle ..." in 📗 Mathematics if the answers seem to be not correct or there's no answer. Try a smart search to find answers to similar questions.
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Polygon Angles - Expii
The sum of a shape's interior (inside) angles will always be 180° × (s - 2), where s represents the number of sides. If you only count one exterior angle at each vertex, the sum of the exterior angles for any polygon will always be 360°.
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Lesson
Lesson 13
13.1: Coordinate Patterns
Plot points in your assigned quadrant and label them with their coordinates.
13.2: Signs of Numbers in Coordinates
Write the coordinates of each point.
\(A=\)
\(B=\)
\(C=\)
\(D=\)
\(E=\)
Answer these questions for each pair of points.
How are the coordinates the same? How are they different?
How far away are they from the y-axis? To the left or to the right of it?
How far away are they from the x-axis? Above or below it?
\(A\) and \(B\)
\(B\) and \(D\)
\(A\) and \(D\)
Pause here for a class discussion.
Point \(F\) has the same coordinates as point \(C\), except its \(y\)-coordinate has the opposite sign.
Plot point \(F\) on the coordinate plane and label it with its coordinates.
How far away are \(F\) and \(C\) from the \(x\)-axis?
What is the distance between \(F\) and \(C\)?
Point \(G\) has the same coordinates as point \(E\), except its \(x\)-coordinate has the opposite sign.
Plot point \(G\) on the coordinate plane and label it with its coordinates.
How far away are \(G\) and \(E\) from the \(y\)-axis?
What is the distance between \(G\) and \(E\)?
Point \(H\) has the same coordinates as point \(B\), except both of its coordinates have the opposite signs. In which quadrant is point \(H\)?
13.3: Finding Distances on a Coordinate Plane
Label each point with its coordinates.
Find the distance between each of the following pairs of points.
Point \(B\) and \(C\)
Point \(D\) and \(B\)
Point \(D\) and \(E\)
Which of the points are 5 units from \((\text-1.5, \text-3)\)?
Which of the points are 2 units from \((\text0.5, \text-4.5)\)?
Plot a point that is both 2.5 units from \(A\) and 9 units from \(E\). Label that point \(F\) and write down its coordinates.
Priya says, "There are exactly four points that are 3 units away from \((\text- 5, 0)\)." Lin says, "I think there are a whole bunch of points that are 3 units away from \((\text- 5, 0)\)."
Do you agree with either of them? Explain your reasoning.
13.4: Plotting Polygons
Here are the coordinates for four polygons. Move the slider to choose the polygon you want to plot. Move the points, in order, to their locations on the coordinate plane. Sketch each one before changing the slider.
Summary
The points \(A = (5, 2), \,B = (\text-5, 2), \, C = (\text-5, \text-2)\), and \(D=(5, \text-2)\) are shown in the plane. Notice that they all have almost the same coordinates, except the signs are different. They are all the same distance from each axis but are in different quadrants.
Notice that the vertical distance between points \(A\) and \(D\) is 4 units, because point \(A\) is 2 units above the horizontal axis and point \(D\) is 2 units below the horizontal axis. The horizontal distance between points \(A\) and \(B\) is 10 units, because point \(B\) is 5 units to the left of the vertical axis and point \(A\) is 5 units to the right of the vertical axis.
We can always tell which quadrant a point is located in by the signs of its coordinates.
\(x\)
\(y\)
quadrant
positive
positive
I
negative
positive
II
negative
negative
III
positive
negative
IV
In general:
If two points have \(x\)-coordinates that are opposites (like 5 and -5), they are the same distance away from the vertical axis, but one is to the left and the other to the right.
If two points have \(y\)-coordinates that are opposites (like 2 and -2), they are the same distance away from the horizontal axis, but one is above and the other below.
When two points have the same value for the first or second coordinate, we can find the distance between them by subtracting the coordinates that are different.
For example, we can find the perimeter of this polygon by finding the sum of its side lengths. Starting from \((\text-2, 2)\) and moving clockwise, we can see that the lengths of the segments are 6, 3, 3, 3, 3, and 6 units. The perimeter is therefore 24 units.
In general:
If two points have the same \(x\)-coordinate, they will be on the same vertical line, and we can find the distance between them.
If two points have the same \(y\)-coordinate, they will be on the same horizontal line, and we can find the distance between them.
Glossary Entries
quadrant
The coordinate plane is divided into 4 regions called quadrants. The quadrants are numbered using Roman numerals, starting in the top right corner
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What is the meaning of axonometric drawing?
What is the meaning of axonometric drawing?
: being or prepared by the projection of objects on the drawing surface so that they appear inclined with three sides showing and with horizontal and vertical distances drawn to scale but diagonal and curved lines distorted an axonometric drawing.
What characterizes a Dimetric drawing?
Dimetric projection is defined as a way of drawing an object so that one axis has a different scale than the other two axis in the drawing. An example of dimetric projection is a technical drawing that shows a 3-dimensional cube with one side of the cube smaller in proportion to the other two sides. noun.
Orthographic Projection: The advantages of using orthographic are: They can show hidden detail and all connecting parts. They can show all dimensions necessary for manufacture. They can be annotated to display material and finishes.
What are the three types of axonometric drawing?
The three types of axonometric projection are isometric projection, dimetric projection, and trimetric projection, depending on the exact angle at which the view deviates from the orthogonal. Typically in axonometric drawing, as in other types of pictorials, one axis of space is shown to be vertical.
What are the three types of oblique drawing?
Oblique drawing
OBLIQUE DRAWING.
OBLIQUE DRAWING : In an oblique drawing the front view is drawn true size, and the receding surfaces are drawn on an angle to give it a pictorial appearance.
Isometric drawings provide a systematic way to draw 3-dimensional objects. Isometric drawings include three axes: one vertical axis and two horizontal axes that are drawn at 30 degree angles from their true position.
What is the difference between isometric projection and Dimetric projection?
And in this corner: an isometric projection is a type of axonometric projection where the same scale is used for each axis and thus it is the most commonly used drawing type. In a dimetric projection only two axes use the same scale while the third (usually the vertical axis) is determined separately.
What are the three types of axonometric projection?
The three types of axonometric projection are isometric projection, dimetric projection, and trimetric projection, depending on the exact angle at which the view deviates from the orthogonal.
What is a first angle projection?
In the first angle projection, the object is placed in the 1st quadrant. The object is positioned at the front of a vertical plane and top of the horizontal plane. First angle projection is widely used in India and European countries. The object is placed between the observer and projection planes.
Who uses orthographic drawing?
An orthographic drawing is a clear, detailed way to represent the image of an object. It may be used by engineers, designers, architects, and technical artists to help a manufacturer understand the specifics of a product that needs to be created.
How is the direction of viewing determined in dimetric projection?
In dimetric pictorials (for methods see dimetric projection ), the direction of viewing is such that two of the three axes of space appear equally foreshortened, of which the attendant scale and angles of presentation are determined according to the angle of viewing; the scale of the third direction (vertical) is determined separately.
How many words are there for dimetric projection?
You must — there are over 200,000 words in our free online dictionary, but you are looking for one that's only in the Merriam-Webster Unabridged Dictionary. Start your free trial today and get unlimited access to America's largest dictionary, with: Ad free! Join Our Free Trial Now! "Dimetric projection."
Which is the best definition of trimetric projection?
trimetric projection in American. a type of axonometric projection in which the object is shown with all three of its principal axes tilted unequally from the plane of viewing.
What is the difference between axonometric projection and orthographic projection?
Axonometry should not be confused with axonometric projection, as in English literature the latter usually refers only to a specific class of pictorials (see below). The orthographic projection is derived from the principles of descriptive geometry and is a two-dimensional representation of a three-dimensional object.
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Common tangents
Two tangent circles and a third circle between the centers. Show that the 3 circles have a common tangent line, and that the middle point of tangency is directly above the point where blue and green touch.
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In Euclidean space, all curves of constant curvature are either straight lines (geodesics) or circles, but in a hyperbolic space of sectional curvature−1,{\displaystyle -1,} the curves of constant curvature come in four types: geodesics with curvature κ=0,{\displaystyle \kappa =0,}hypercycles with curvature 0<|κ|<1,{\displaystyle 0<|\kappa |<1,} horocycles with curvature |κ|=1,{\displaystyle |\kappa |=1,} and circles with curvature |κ|>1.{\displaystyle |\kappa |>1.}
Any two horocycles are congruent, and can be superimposed by an isometry (translation and rotation) of the hyperbolic plane.
A horocycle can also be described as the limit of the circles that share a tangent at a given point, as their radii tend to infinity, or as the limit of hypercycles tangent at the point as the distances from their axes tends to infinity.
Two horocycles with the same centre are called concentric. As for concentric circles, any geodesic perpendicular to a horocycle is also perpendicular to every concentric horocycle.
longer than the length of the arc of a hypercycle between those two points and
shorter than the length of any circle arc between those two points.
The distance from a horocycle to its centre is infinite, and while in some models of hyperbolic geometry it looks like the two "ends" of a horocycle get closer and closer together and closer to its centre, this is not true; the two "ends" of a horocycle get further and further away from each other.
A regular apeirogon is circumscribed by either a horocycle or a hypercycle.
If C is the centre of a horocycle and A and B are points on the horocycle then the angles CAB and CBA are equal.[1]
The area of a sector of a horocycle (the area between two radii and the horocycle) is finite.[2]
Standardized Gaussian curvature
The lengths of an arc of a horocycle between two points is: s=2sinh(12d)=2(coshd−1){\displaystyle s=2\sinh \left({\frac {1}{2}}d\right)={\sqrt {2(\cosh d-1)}}} where d is the distance between the two points, and sinh and cosh are hyperbolic functions.[3]
The length of an arc of a horocycle such that the tangent at one extremity is limiting parallel to the radius through the other extremity is 1.[4] the area enclosed between this horocycle and the radii is 1.[5]
The ratio of the arc lengths between two radii of two concentric horocycles where the horocycles are a distance 1 apart is e : 1.[6]
Representations in models of hyperbolic geometry
Poincaré disk model
In the Poincaré disk model of the hyperbolic plane, horocycles are represented by circles tangent to the boundary circle; the centre of the horocycle is the ideal point where the horocycle touches the boundary circle.
Metric
Horocycle flow
Every horocycle is the orbit of a unipotent subgroup of PSL(2,R) in the hyperbolic plane. Moreover, the displacement at unit speed along the horocycle tangent to a given unit tangent vector induces a flow on the unit tangent bundle of the hyperbolic plane. This flow is called the horocycle flow of the hyperbolic plane.
Identifying the unit tangent bundle with the group PSL(2,R), the horocycle flow is given by the right-action of the unipotent subgroup U={ut,t∈R}{\displaystyle U=\{u_{t},\,t\in \mathbb {R} \}}, where:
ut=±(1t01).{\displaystyle u_{t}=\pm \left({\begin{array}{cc}1&t\\0&1\end{array}}\right).}
That is, the flow at time t{\displaystyle t} starting from a vector represented by g∈PSL2(R){\displaystyle g\in \mathrm {PSL} _{2}(\mathbb {R} )} is equal to gut{\displaystyle gu_{t}}.
If S{\displaystyle S} is a hyperbolic surface its unit tangent bundle also supports a horocycle flow. If S{\displaystyle S} is uniformised as S=Γ∖H2{\displaystyle S=\Gamma \backslash \mathbb {H} ^{2}} the unit tangent bundle is identified with Γ∖PSL2(R){\displaystyle \Gamma \backslash \mathrm {PSL} _{2}(\mathbb {R} )} and the flow starting at Γg{\displaystyle \Gamma g} is given by t↦Γgut{\displaystyle t\mapsto \Gamma gu_{t}}. When S{\displaystyle S} is compact, or more generally when Γ{\displaystyle \Gamma } is a lattice, this flow is ergodic (with respect to the normalised Liouville measure). Moreover, in this setting Ratner's theorems describe very precisely the possible closures for its orbits.[7]
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The core concepts of trigonometry are developed from a circle with radius equal to \(1\) unit, drawn in the \(xy\)-coordinate plane, centered at the origin. This circle is given a name: the unit circle (Figure \(7.1.1\) below). Just like a \(12\)-hour clock with values of time from \(1\) to \(12\), trigonometric functions are periodic, meaning the same values are reproduced with each \(360˚\) revolution.
An angle is in standard position (see Figure \(7.1.2\) above) if its initial side is along the positive \(x\)-axis and its vertex is at the origin: point \((0,0)\). The following angles are in standard position. An angle that rotates in the counter-clockwise direction is a positive angle. An angle that rotates in the clockwise direction is a negative angle.
Coterminal Angles
Two or more standard angles that share common terminal sides are said to be coterminal angles. For example, \(30˚\) and \(390˚\) are coterminal angles.
More formally: Every angle \(B\) is coterminal with angle \(A\) where \(B = A + 360˚k\), \(k =\) any integer.
Example 7.1.1
The expression \(315˚ + 360˚k\) gives the angles coterminal with \(315˚\). State the coterminal angles the equation generates using the following \(k\)-values: \(k = −2, −1, 1, 2\). Then sketch the angles.
Solution
We substitute the given \(k\)-values into the expression \(315˚ + 360˚k\).
\(k = -2\)
\(k = -1\)
\(k = 1\)
\(k = 2\)
\(315˚ + 360˚(\textcolor{red}{-2}) = -405˚\)
\(315˚ + 360˚(\textcolor{red}{-1}) = -45˚\)
\(315˚ + 360˚(\textcolor{red}{1}) = -675˚\)
\(315˚ + 360˚(\textcolor{red}{2}) = 1035˚\)
Below, the angles are sketched. Do you see that each angle shares the same terminal side? All four angles are coterminal with \(315˚\), and coterminal with each other.
Circle Centered at the Origin
Every ordered pair \((x, y)\) on a circle is associated with a right triangle. The right triangle has horizontal distance \(x\), vertical distance \(y\), and hypotenuse = radius = \(r\).
Equation of a Circle
The equation of a circle of radius \(r\) centered at the origin:
\[x^2 + y^2 = r^2\]
Note: the \(x\)-coordinate and the \(y\)-coordinate can take on negative values, depending on the quadrant of the terminal side of the angle.
Example 7.1.2
Find the \(y\)-coordinate of point A \(\left(−\dfrac{5}{9} , y \right)\) if point A lies in QIII on the unit circle.
Solution
The unit circle has radius \(r = 1\). Trigonometry weds algebra and geometry with visual sketches. Create a sketch before jumping into a solution. This helps you see the answer.
Special Right Triangles
Using the Pythagorean Theorem, one can drive the following two templates for special right triangles: \(45˚\)-\(45˚\)-\(90˚\) triangles and \(30˚\)-\(60˚\)-\(90˚\) triangles.
Since right triangles and circles are inextricably tied to each other, the acute angles \(30˚\), \(45˚\), \(60˚\) are frequently useful for finding exact values in trigonometry; these solutions do not require the use of a calculator.
Examples of special right triangles and their solutions, can be viewed in these videos:
Try It! (Exercises)
For #1-5, draw the angle \(\theta\) in standard position.
\(\theta = 210˚\)
\(\theta = −300˚\)
\(\theta = 150˚\)
\(\theta = 270˚\)
\(\theta = −135˚\)
For #6-10, state any two angles coterminal with the given angle \(\theta\). Give one positive coterminal angle and one negative coterminal angle.
\(\theta = 30˚\)
\(\theta = 60˚\)
\(\theta = 90˚\)
\(\theta = 180˚\)
\(\theta = 240˚\)
For #11-15, state the equation of the circle with the given radius.
\(r = 4\)
\(r = \dfrac{1}{2}\)
\(r = 3 \sqrt{2}\)
\(r = \dfrac{\sqrt{6}}{2}\)
\(r = \sqrt{\dfrac{101}{62}}\)
Fill in the blanks: A unit circle is a circle with \(\underline{\;\;\;\;\;\;\;\;\;\;}\) equal to one unit. The circle is centered at \(\underline{\;\;\;\;\;\;\;\;\;\;}\). The circle has equation: \(\underline{\;\;\;\;\;\;\;\;\;\;}\).
For #17-24, The given point lies on a circle with given radius and terminal side in the given quadrant. Find the missing coordinate of the given ordered pair.
Point on Circle
Radius
Quadrant in which \(\theta\) terminates.
17. \((2, y)\)
\(r = 5\)
\(\theta ∈\) QI
18. \((x, \sqrt{6})\)
\(r = 3\)
\(\theta ∈\) QII
19. \((−3\sqrt{7}, y)\)
\(r = \sqrt{77}\)
\(\theta ∈\) QIII
20. \((x, −4\sqrt{5})\)
\(r = 3\sqrt{15}\)
\(\theta ∈\) QIV
21. \(\left( \dfrac{3}{4} , y \right)\)
\(r=1\)
\(\theta ∈\) QI
22. \(\left(x, \dfrac{3}{16} \right)\)
\(r=1\)
\(\theta ∈\) QII
23. \(\left(− \dfrac{\sqrt{5}}{4} , y \right)\)
\(r=1\)
\(\theta ∈\) QIII
24. \(\left(x, − \dfrac{3\sqrt{2}}{8} \right)\)
\(r=1\)
\(\theta ∈\) QIV
For #25-29, find the fraction of a full revolution. Then sketch the angle.
\(\dfrac{1}{4} \cdot 360˚\)
\(\dfrac{1}{2} \cdot 360˚\)
\(\dfrac{3}{4} \cdot 360˚\)
\(\dfrac{1}{8} \cdot 360˚\)
\(\dfrac{1}{6} \cdot 360˚\)
For #30-35, solve for the \(2\) missing side lengths. Give exact answers. No calculators.
For #36-38, Find the coordinates of the ordered pair \((x, y)\) on the unit circle with the given standard angle. Use special right triangle relationships. Give exact values of \(x\) and \(y\).
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CBSE Maths Notes For Class 9
CBSE Class 9 Maths Notes is an interesting resource that focuses on the study of numbers, shapes and patterns. Besides, maths help to solve problems that can be related to the real world, and as everything in this universe follows a pattern, the subject of mathematics can be applied in many instances. Students can understand these concepts effortlessly with the help of the CBSE Notes given at BYJU'S.
Students can access CBSE Class 9 notes of maths by clicking on the links provided in the table given below.
Studying mathematics can be a nightmare for some students as it contains a lot of complex concepts, formulas, and sums, and it basically requires a lot of critical thinking. But again, there are some students who really enjoy solving problems in mathematics. These CBSE Class 9 notes are prepared by referring to CBSE Class 9 syllabus.
With an aim to make math easy, we at BYJU'S are providing a comprehensive set of CBSE Class 9 Maths notes which will help students to study effectively as well as score well in the examination. These maths notes have been prepared by experts for CBSE Class 9, who have further taken great care to use simple language and style so that students are able to understand concepts easily and quickly. These notes act as a great reference tool and will help students have quick revisions of all the chapters before exams.
Frequently Asked Questions on CBSE Class 9 Maths Notes
Q1
What is 'Heron's formula'?
The area of a triangle can be determined using Heron's Formula = A = √{s(s-a)(s-b)(s-c)}.
Q2
What is 'Euclid's geometry'?
Euclidean geometry is the study of plane figures and solid figures on the basis of axioms and theorems employed by the Greek mathematician Euclid.
Q3
What are quadrilaterals?
A quadrilateral can be defined as a closed, two-dimensional shape which has four straight sides.
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In a triangle, one of the sides is 10, the other 10√2, and the angle between them is 45 °. Find the area of a triangle.
Knowing the two sides of the triangle and the angle between them, you can find the area of the triangle by the formula S tr. = 1/2 * ab * sin A, where a and b are known sides, and angle A is the angle between sides a and b
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Elements of Geometry: Containing the First Six Books of Euclid, with a ...
RIGONOMETRY is the application of Arithmetic to Geometry: or, more precisely, it is the application of number to express the relations of the sides and angles of triangles to one another. It therefore necessarily supposes the elementary operations of arithmetic to be understood, and it borrows from that science several of the signs or characters which peculiarly belong to it. Thus, the product of two numbers A and B, is either denoted by A.B or AXB; and the products of two or more into one, or into more than one, as of A+B into C, or of A+B into C+D, are expressed thus: (A+B) C, (A+B) (C+D), or sometimes thus, A+B × C, and A+B × C+D. The quotient of one number A, divided by another B, is written
The sign is used to signify the square root: Thus M is the square root of M, or it is a number which, if multiplied into itself, will produce M. So also, ✓ M2 + N2 is the square root of M2+N2, &c. The elements of Plane Trigonometry, as laid down here, are divided into three sections; the first explains the principles; the second delivers the rules of calculation; the third contains the construction of trigonometrical tables, together with the investigation of some theorems, useful for extending trigonometry to the solution of the more difficult problems.
SECTION I.
LEMMA I.
An angle at the centre of a circle is to four right angles as the arch on which it stands is to the whole circumference.
Let ABC be an angle at the centre of the circle ACF, standing on the circumference AC: the angle ABC is to four right angles as the arch AC to the whole circumference ACF.
Produce AB till it meet the circle in E, and draw DBF perpendicular to AE.
Then, because ABC, ABD are two angles at the centre of the circle, ACF, the angle ABC is to the angle ABD as the arch AC to the arch AD, (33. 6.); and therefore also, the angle ABC is to four times the angle ABD as the arch AC to four times the arch AD (4. 5.).
But ABD is a right angle, and therefore four times the arch AD is equal to the whole circumference ACF; therefore, the angle ABC is
E
H
K
G
to four right angles as the arch AC to the whole circumference ACF. COR. Equal angles at the centres of different circles stand on arches which have the same ratio to their circumferences. For, if the angle ABC, at the centre of the circles ACE, GHK, stand on the arches AC, GH, AC is to the whole circumference of the circle ACE, as the angle ABC to four right angles; and the arch HG is to the whole circumference of the circle GHK in the same ratio. There fore, &c.
DEFINITIONS. I.
If two straight lines intersect one another in the centre of a circle, the arch of the circumference intercepted between them is called the Measure of the angle which they contain. Thus the arch AC is the measure of the angle ABC.
II.
If the circumference of a circle be divided into 360 equal parts, each of these parts is called a Degree; and if a degree be divided into 60 equal parts, each of these is called a minute; and if a Minute be divided into 60 equal parts, each of them is called a Second, and so And as many degrees, minutes, seconds, &c. as are in any arch, so many degrees, minutes, seconds, &c. are said to be in the angle measured by that arch.
on.
COR. 1. Any arch is to the whole circumference of which it is a part, as the number of degrees, and parts of a degree contained in it is to the number 360. And any angle is to four right angles as the number of degrees and parts of a degree in the arch, which is the measure of that angle, is to 360.
COR. 2. Hence also, the arches which measure the same angle, whatever be the radii with which they are described, contain the same number of degrees, and parts of a degree. For the number of degrees and parts of a degree contained in each of these arches has the same ratio to the number 360, that the angle which they measure has to four right angles (Cor. Lem. 1.).
The degrees, minutes, seconds, &c. contained in any arch or angle, are usually written as in this example, 49°. 36′. 24". 42′′; that is, 49 degrees, 36 minutes, 24 seconds, and 42 thirds.
III.
Two angles, which are together equal to two right angles, or two arches which are together equal to a semicircle, are called the Supplements of one another.
IV.
A straight line CD drawn through C, one of the extremities, of the arch AC, perpendicular to the di
ameter passing through the other extremity A, is called the Sine of the arch AC, or of the angle ABC, of which AC is the measure.
COR. 1. The sine of a quadrant, or of a right angle, is equal to the radius.
COR. 2. The sine of an arch is half the chord of twice that arch: this is evident by producing the sine of any arch till it cut the circumference.
V.
I
H
H
K
The segment DA of the diameter passing through A, one extremity of the arch AC, between the sine CD and the point A, is called the Versed sine of the arch AC, or of the angle ABC.
VI.
A straight line AE touching the circle at A, one extremity of the arch AC, and meeting the diameter BC, which passes through C the other extremity, is called the Tangent of the arch AC, or of the angle ABC.
COR. The tangent of half a right angle is equal to the radius.
VII.
The straight line BE, between the centre and the extremity of the tangent AE is called the Secant of the arch AC, or of the angle ABC. COR. to Def. 4, 6, 7, the sine, tangent, and secant of any angle ABC, are likewise the sine, tangent, and secant of its supplement CBF.
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Turtle objects have methods and attributes. For example, a turtle has a position and when you move the turtle forward, the position changes. Think about the other methods shown in the summary above. Which attibutes, if any, does each method relate to? Does the method change the attribute?
Use for loops to make a turtle draw these regular polygons
(regular means all sides the same lengths, all angles the same):
An equilateral triangle
A square
A hexagon (six sides)
An octagon (eight sides)
# draw an equilateral triangleimportturtlewn=turtle.Screen()norvig=turtle.Turtle()foriinrange(3):norvig.forward(100)# the angle of each vertice of a regular polygon# is 360 divided by the number of sidesnorvig.left(360/3)wn.exitonclick()
Write a program that asks the user for the number of sides, the length of the side, the color, and the fill color of a
regular polygon. The program should draw the polygon and then fill it in.
A drunk pirate makes a random turn and then takes 100 steps forward, makes another random turn, takes another 100 steps, turns another random amount, etc. A social science student records the angle of each turn before the next 100 steps are taken. Her experimental data is 160,-43,270,-97,-43,200,-940,17,-86. (Positive angles are counter-clockwise.) Use a turtle to draw the path taken by our drunk friend. After the pirate is done walking, print the current heading.
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Home » UsingSteps for construction:
i) Draw a line segment AB = 6 cm
ii) Draw a ray at A, making an angle of 60owith BC.
iii) With B as centre and radius = 6.2 cm draw an arc which intersects AX ray at C.
iv) Join BC.
Then, ΔABC is the required triangle.
v) Draw the perpendicular bisectors of AB and AC intersecting each other at O.
vi) With centre O, and radius as OA or OB or OC, draw a circle which will pass through A, B and C.
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What is volume of triangle which has no right angle amongthree angles?
A triangle is a 2D shape, so as such it has no volume. (Since it
is a length and a width, but has no depth).
The equivalent of a triangle in 3D is a triangular based
pyramid, which clearly does have a volume.
You may have meant "area" rather than volume?
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Geometry
Geometry is the study of different types of shapes, figures and sizes in Maths or in real life. In geometry, we learn about different angles, transformations and similarities in the figures.
Plane Geometry deals with flat shapes which can be drawn on a piece of paper. These include lines, circles & triangles of two dimensions. Plane geometry is also known as two-dimensional geometry.
All the two-dimensional figures have only two measures such as length and breadth. It does not deal with the depth of the shapes. Some examples of plane figures are square, triangle, rectangle, circle, and so on.
The important terminologies in plane geometry are:
Point
Line
Angles
Point
A point is a precise location or place on a plane. A dot usually represents them. It is important to understand that a point is not a thing, but a place. Also, note that a point has no dimension; preferably, it has the only position.
Line
The line is straight (no curves), having no thickness and extends in both directions without end (infinitely). It is important to note that it is the combination of infinite points together to form a line. In geometry, we have a horizontal line and vertical line which are x-axis and y-axis respectively.
Line Segment – If a line has a starting and an endpoint then it is called a Line Segment.
Ray – If a line has a starting point and has no endpoint is called Ray.
Incidence Axioms on Lines • A line contains infinitely many points • An infinite number of lines can be drawn to pass through a given point • One and only one line can be drawn to pass through two given points A and B.
Collinear Points Three or more points are said to be collinear, if there is a line which contains them all. In the above figure; P, Q, R are collinear points.
INTERSECTING LINES Two lines having a common point are called intersecting lines. The point common to two given lines is called their point of intersection. In the figure, the lines AB and CD intersect at a point O.
PARALLEL LINES Two lines l and m in a plane are said to be parallel, if they have no point in common and is written as l || m. The distance between two parallel lines always remains the same.
CURVES Curves can be defined as figures that flow smoothly without a break. A line is also a curve, and is called a straight curve. Simple curves Curves that do not intersect themselves are called simple curves. Open curves Curves whose end points do not meet are called open curves. Closed curves Curves whose end points join to enclose an area are called closed curves. For a closed curve, we can identify three regions: The interior of the curve: These points are in the interior of the closed curve. Boundary of the curve: These points are on the boundary of the closed curve. Exterior of the curve: These points are in the exterior of the closed curve. The interior of a curve together with its boundary is called its "region".
Angles in Geometry
An angle is made up of two rays starting from a common end point. In this figure BA and BC rays have one common end point, that is, B. The rays BA and BC are called the arms or sides of the angle. The common end point B is the vertex of the angle. We name the above angle as ∠BAC.
Types of Angle
Acute Angle – An Acute angle (or Sharp angle) is an angle smaller than a right angle ie. it can range between 0 – 90 degrees.
Obtuse Angle – An Obtuse angle is more than 90 degrees but is less than 180 degrees.
Right Angle – An angle of 90 degrees.
Straight Angle – An angle of 180 degrees is a straight angle, i.e. the angle formed by a straight line
Polygons in Geometry
A plane figure that is bounded by a finite chain of straight line segments closing in a loop to form a closed polygonal chain or circuit.
The name 'poly' refers to multiple. An n-gon is a polygon with n sides; for example, a triangle is a 3-gon polygon.
General Formula for Sum of internal Angles of a polygon – (n - 2) x 180
Quadrilateral : A 4-sided polygon with four edges and four vertices. Sum of internal angles is 360 degrees.
Types Quadrilateral :
Square – Has 4 equal sides and vertices which are at right angles.
Rectangle – Has equal opposite sides and all angles are at right angles.
Parallelogram – has two pairs of parallel sides. The opposite sides & opposite angles are equal in measure.
Rhombus – Has all the four sides to be of equal length. However, they do not have its internal angle to be 90 degrees
Trapezium – Has one pair of opposite sides to be parallel.
Pentagon : A plane figure with five straight sides and five angles
Hexagon : A plane figure with six straight sides and six angles
Heptagon : A plane figure with seven sides and seven angles
Octagon : A plane figure with eight straight sides and eight angles.
Nonagon : A plane figure with nine straight sides and nine angles.
Decagon : A plane figure with ten straight sides and ten angles.
Circle in Geometry
A circle is formed by a point moving at the same distance from a fixed point. The fixed point is the centre of the circle. A circle is also a simple closed curve however, it does not have any sides or angles. The fixed point O is the centre of the circle. The fixed distance OP = OQ is the radius of the circle. The distance around the circle is its circumference.
Circumference The line that forms the boundary of a circle is called its circumference. The part enclosed by the circumference of a circle is called the interior of the circle. The part left outside the circle is said to be the exterior of the circle. Some points may lie on the circumference of the circle. Radius A line segment that joins the centre of the circle and a point on the circumference is called the radius of the circle. The radius of a circle is half of the diameter. Chord A chord is a line segment joining two points that lie on a circle. Diameter A chord passing through the centre of the circle is called its diameter. A diameter is the longest chord of a circle. Arc An arc is a part of the circumference of a circle. Sector The part of a circle enclosed by two radii and an arc is called a sector. Segment The part of a circle that is enclosed by a chord and an arc is called a segment of the circle. Semi-circle A diameter of a circle divides it into two halves. Each half is called a semi-circle.
Similarity and Congruency in Geometry
Similarity – Two figures are said to be similar if they have the same shape or have an equal angle but do not have the same size.
Congruence – Two figures are said to be Congruent if they have the same shape and size. Thus, they are totally equal.
Solid Geometry (Three-dimensional geometry)
Solid Geometry deals with 3-dimensional objects like cubes, prisms, cylinders & spheres. It deals with three dimensions of the figure such as length, breadth and height. But some solid solids do not have faces (e.g. sphere).
Solid geometry is the study of three dimensions in Euclidean space. The objects which are around us are three-dimensional. All the three-dimensional shapes are obtained from the rotation operation of 2D shapes. The important attributes of 3D shapes are:
Faces
Edges
Vertices
Edges
An edge is defined as the line segment on the boundary that joins one vertex to the other vertex. It means that it joins one corner point to the other. It forms the skeleton of 3D shapes. In other words, it can be defined as the faces, that meet in the straight line is called edge. Following are the list of edges for the different solid shapes:
Triangular Prism : 9
Cube : 12
Rectangular prism : 12
Pentagonal Prism : 15
Hexagonal Prism : 18
Triangular Pyramid : 6
Square Pyramid : 8
Pentagonal Pyramid : 10
Hexagonal Pyramid : 12
Faces
We know that all the geometric shapes are made up of flat surface called faces. It is a flat surface enclosed by the edges. For any three-dimensional shapes, the face should be a two-dimensional figure. The list of the number of faces for different solid shapes are given below:
Triangular Prism : 5
Cube : 6
Rectangular prism : 6
Pentagonal Prism : 7
Hexagonal Prism : 8
Triangular Pyramid : 4
Square Pyramid : 5
Pentagonal Pyramid : 6
Hexagonal Pyramid : 7
Vertices
A vertex is defined as the point where the edges of the solid figure meet at each other. In other words, it can be said that, the point where the adjacent sides of the polygon meet. The vertex is the corners where the edges meet. The number of vertices for different solid shapes in geometry is as follows:
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The figure shows an equilateral triangle ABC in which all the dimensions are given in centimetres. a) Find the value of x. b) Use any two sides and an included angle or otherwise to calculate the area of triangle ABC.
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To draw triangles with specific angles.
Is there any tool by which,one can draw an triangle with specific angles?say i want one angle of triangle to be 5 degrees and others would be obviously 90degrees and 85degrees in case of right angled triangle.But the basic problem is how can you draw or set the one angle of triangle to be 5 degrees?
If there is not any tool available,is there any alternative solution to this problem?My basic aim is to draw triangles with different angles starting 5degrees onwards!
This code will generate a triangle with the given three points in the current cell on the layer currently selected.
You can compute the points using any formula you wish. You can also put the triangle generation into a loop and thus generate a whole array of shapes with a variation of parameters.
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The Lego robot shows an illustrative example of the area of squares in
Pythagoras' theorem.
Goals:
Motivating the students – revision of Pythagorean theorem, calculating the interior and exterior angles, two-dimensional demonstration of the ratio of areas. If a pen is used, it draws a curve of the route. Pay attention to the repetition of moves.
Learning programming in Python. Students have to analyse the necessary moves, indicate programming steps, take positive and negative rotation into account, and think about when the pen needs to be raised or lowered. The lengths of catheti need to be adapted to the performance and size of the Lego robot.
The LEGO Mindstorm EV3 Robot that coincides with this tutorial comes from building specific sections found in the LEGO Mindstorm Education Core Set building instructions. You will need to build the main body for the robot or you could prepare your own robot - watch below (compare modul ,,Robot drawing triangles")
In a right triangle, a cathetus (plural: catheti), commonly known as a
leg, is either of the sides that are adjacent to the right angle. It is
occasionally called a "side about the right angle". The side opposite
the right angle is the hypotenuse.
Students enter the data of catheti a and b. The other necessary data are
calculated by the programme, taking into account:
The length of the hypotenuse c – to calculate the length of hypotenuse use \(a^{2} + b^{2} = c^{2}\)
#!/usr/bin/env python3
calculating the necessary values
c=math.sqrt(a*a+b*b) //we calculate the hypotenuse
beta=math.atan(a/b) //we calculate beta angle (currently in radian measure)
PI=math.pi //we convert pi into a constant
We make our work easier by using the following functions:
def forward(length): //we create a function with a parameter (length)
to move forward
tank_drive.on_for_rotations(SpeedPercent(50), SpeedPercent(50), length)
def rotate(angle,orientation): // we create a function with two parameters (angle and orientation) to rotate through a certain angle in + or – direction
tank_drive.on_for_rotations(SpeedPercent(-orientation*50), SpeedPercent(orientation*50), angle)
Students write a programme on the basis of programme steps they have
planned.
Comment: Angles alpha, beta and gamma and lengths a, b and c have to be
multiplied by empirically defined factor in order for the rotation to be
exactly 90 degrees (not more, not less). The same needs to be taken into
account when we plan moves that make the robot return to the initial
position.
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is inscribed in an isosceles trapezoid with bases 8 and 18
[#permalink]
25 Jul 2023, 22:55
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Kudos 01:45
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Expert Reply
vedha0 wrote: 09:52
A circle is inscribed in an isosceles trapezoid with bases 8 and 18 as shown above. What is the area of the circle?
A. \(25\pi\) B. \(30\pi\) C. \(36\pi\) D. \(49\pi\) E. \(64\pi\) Area of circle = \(r^2\pi\) Therefore the answer choice must be a multiple of square of a number(most likely an integer as we see the choice available). Here itself we can reject B as 30 is not a square of an integer.
Since this is an isosceles trapezoid, the sides non parallel sides are equal. The figure is such that circles fits in such a manner that all sides touch circle uniformly i.e. point of touch divides parallel sides in half and non parallel ones in equal ratios. the smaller part of those non parallel sides is of the length 4(8/2) and bigger one is of the length 9(18/2). Now that we know the length of the non parallel sides is 13(4 + 9) we can get the distance between the tow parallel sides using Pythagoras theorem, which gives us 12. Thus, radius is 6.
Now if we see the answer choices we have only one choice that is a multiple of 6 or 6ˆ2.
Note: the non parallel sides would touch circle such that the point of touch divides them in the ratio of length of parallel sides i.e. 8/18 = 4/9. Actually, here it is 4 and 9.
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Trigonometry - Graph of Tan lesson
Hi, I've really enjoyed using Khan Academy and I have learned a ton of trig very rapidly (for me). I've found each lesson easy to follow and clear, and I also like that I can pause and guess where things are going, or watch each concept get developed step-by-step if I can't predict.
The exception is the lesson where the tangent of theta is graphed. Sine and cosine and their intersection points were all great, but this video jumps right to the fact that this is a periodic function reaching infinity every pi radians, all at once, without letting us see that concept develop. First pi/4's tangent is graphed, then negative pi/4, and it looks like a straight line is forming, but we don't test further to see for ourselves that's not what's happening--we are simply told, and the graph is drawn as a curve without our understanding why. Please fill in the gaps hear and lead us to the solution, like you usually do. Thank you!
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Let ABC be the triangle and P the given point without. Draw PE parallel to AB, cutting AC in D. Make parallelogram DEFA = given square. On F erect the perpendicular FG-PD, and make GQ-PE. Connect P with Q, then will PQ be the required line.
31; in Beшan and Smith's translatión, p. 34.
II. Solution reported by the PROPOSER.
The elegant solution by Lill (reported without proof by d'Ocagne at the Second International Congress of Mathematicians, Paris, 1900) is so simple that
Y Q
B
A
OM
NH 2
the Proposer has used it in his courses in elementary algebra. For the graphic solution of x2+px+q=0, choose two perpendicular lines Ox and Oy, lay off unit length OA on Oy, length OH on Ox containing -p units (to right or left of O, according asp is or ), length HB on parallel to Oy containing q units. If the circle on AB as diameter cuts Ox at M and N, then OM and ON, on the same scale, are the required roots. In proof, let Q be the second point of intersection of the circle Oy, then OQ=HB, since OHBQ is a rectangle; OM
Let ax + bx+c=0, be the general quadratic. On the line AD, lay off AB-2 units, and BD=c/2a. On AD as a diameter describe the circle AED. At B erect the perpendicular BE. With E as a center and a radius equal to b/2a, describe an are intersecting AD, or, AD produced, in C. Then with C as a center and a radius equal to CB describe the circle FBG intersecting EC in G and F in the order.
E, G, C, F. Then EG is equal to the value of one root of the quadratic and EF is equal to the other. For
This construction only gives the absolute values of the real roots. responding algebraic values must be assigned.
If b2/4a2=c/a, then CE-BE, which equals x. If b2/4a2 <c/a, the construction is impossible.
IV. Solution by A. H. HOLMES, Brunswick, Maine.
The corres
Describe a circle of radius a, and from a point on the circumference A draw tangent AB/b. Then from B draw through the center of the circle O the line BD cutting the circumference in C and D. By the principles of plane geometry the lines CB and BD will represent the unknown quantity in the quadratic equation x2+2ax=b.*
V. Solution by L. LELAND LOCKE, Brooklyn, N. Y.
We have four cases to consider. These may be reduced to two.
(3),
(4).
Let q=r.s, r and s being any suitable factors of q.
Case I. Equations 1, 2. Draw a circumference 1, and in it a chord AB=p, using any convenient unit of measure. Tangent to this chord and concentric with circle 1 draw circle 2. In circle 1 place a chord MNP such that MP=r and NP-s., Through P draw chord CD tangent to circle 2. PC and PD are the roots of the quadratics. Their values may be found, using the same unit of measure as before.
Proof. AB-CD=p, CP=x,, PD=x-p-x; CP.PD-NP.PM=s.r, x(p-x)=rs,
x-px1+rs-07
2 1
Similarly, x-px+rs=0 equation 1.
*There are four cases, since a and b may each be positive or negative. The above solution suggests at once the following: Construct a circle of radius a; at the distance /b from the diameter AB draw a parallel chord which intersects the circle (if at all) in A'B'. Draw B'H perpendicular to AB; it is now evident that AH, HB are roots of the quadratic (2a-x)x-b; i. e. of x2-2ax+b=0.
In order to solve x2-2ax-b we observe that by changing the signs of both roots the equation 2+2ax =b is obtained, and Mr. Holmes' solution applies.
Similarly, in order to solve x2+2ax+b=0 we observe that its roots are the negatives of the roots of x2-2ax+b=0, and the construction at the beginning of this foot-note applies. ED.
Equation 2, x2-px2+q=0, may be solved in a similar manner by changing the sign of px and proceeding as above. The roots being the same in numerical value but opposite in sign.
Case II. Equations 3, 4. The figure for Case II differs from that of Case I only in that P is outside of the circle 1; the points CPD being now in the order POD. PC, PD-x=x+p, PN.PM-PC.PD; r.s=x(x1+p), x2+px, -rs =0. Similarly, x2+px, -rs-0.
2
1
2 1
1
Equation 4. is solved by changing sign of pr and proceeding as with equation 3, since the roots are the opposite of the roots of equation (3).
In Case I if the roots are imaginary, the point P falls within circumfer
ence 2, and the graphic method fails.
VI. Solution by G. B. M. ZERR. A. M., Ph. D., Parsons. W. Va.
Upon AB describe a semi-circle. Let C be the mid-point of the semi-circular arc. Draw AC, BC. Let G be a point on the line AC; on GC describe a circle center D. Through D draw BFDE. Then taking BE, BF positive; EB, FB negative, these lines are the roots of a quadratic having 2CD for the coefficient of the first power of the unknown quantity, and 1/(2AB) for the absolute term, the coefficient of the second power of the unknown quantity being taken unity..
then for x2+bx-c, EB=-[b+√(b2+4c)], BF=-1[b—√(b2+4c)]. For x2 bx-c, BE [b+1/(b2+4c)], FB={[b−1/(b2+4c)]. If c be negative, the results still hold.
Also solved by G. W. Greenwood, B. A. (Oxon), Professor of Mathematics, and Astronomy in McKendree College, Lebanon, Ill., by use of circle and hyperbola.
219A. Proposed by H. F. MacNEISH, A.B., Assistant in Mathematics, University High School, Chicago, Ill. Draw a line through a given point which shall divide a given quadrilateral into two equivalent parts; (1) when the point lies in a side of the quadrilateral, (2) when the point is without, (3) within the quadrilateral..
..POL+LCD PODL=HCL+LCD-HCD-ABCD. (2). Let R be the point without the quadrilateral.
Draw RK parallel to
FD meeting FC in K.
Join KG and draw CI parallel to KG, join KI. Then
ICG-ICK and FCI=FCI..
..FCG=FKI. Draw IJ parallel to FC, then parallelogram FKJI=FCH. At the point I draw IZ perpendicular to FD and equal to RK, draw ZL-RJ.
:. FPL-FENM-FCH... FPL-FBA=FCH-FCS-SCH. ..ABPL-SCH-ABCD::
Also solved by G. W. Greenwood, B. A. (Oxon).
220. Proposed by G. B.M. ZERR, A. M., Ph. D., Parsons, West Va.
Two triangles are circumscribed to a given triangle ABC, having their sides perpendicular to the sides of the given triangle. Prove that the two triangles are equal, and find the area of these triangles.
The perpendiculars to a side of the given triangle at its extremities, which are corresponding sides of the circumscribed triangles, are symmetrical with respect to any point on the perpendicular bisector of that side. Hence the two triangles are symmetric with respect to the circumcenter of the given triangle and are therefore equal. The area of either is equal to
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71
Page vi ... semicircles on the three sides of an isosceles right - angled triangle , and observing that the sum of the two lunes , between the two quadrants of the larger semicircle and the two smaller semicircles , was equal to the area of the ...
Page 3 ... semicircle is the figure contained by a diameter and the part of the circumference cut off by the diameter . XIX . The centre of a semicircle is the same with that of the circle . XX . Rectilineal figures are those which are contained ...
Page 43 ... semi - circle ; or without the figure , as in certain conic lines . " Def . XXIV - XXIX . Triangles are divided into three classes by reference to the relations of their sides , and into three other classes by reference to their angles ...
Page 66 ... semi- circle BHF , and produce DE to meet the circumference in H. The square described upon EH shall be equal to the given rectilineal figure A. Join GH . Then because the straight line BF is divided into two equal parts in the point G ...
Page 90 ... semicircle is said to be greater than any acute rectilineal angle , and the remaining angle less than any rectilineal angle . " Q.E.D. COR . From this it is manifest , that the straight line which is drawn at right angles to
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Asked by: Velma Seiferlein
sciencespace and astronomy
Is the transformation a rigid motion?
9In this manner, how do you tell if a transformation is a rigid motion?
A rigid motion is a transformation (of the plane) that "preserves distance". In other words, if A is sent/mapped/transformed to A′ and B is sent to B′, then the distance between A and B (the length of segment AB) is the same as the distance between A′ and B′ (the length of segment A′B′).
Likewise, which transformation is a form of rigid motion and slides a figure? translation
People also ask, are all translations considered a rigid motion?
b) the relative position of the points stays the same. There are four types of rigid motions that we will consider: translation , rotation, reflection, and glide reflection. In a translation, everything is moved by the same amount and in the same direction. Every translation has a direction and a distance.
Is a dilation an example of a rigid motion?
A dilation is not considered a rigid motion because it does not preserve the distance between points. Under a dilation where , and , , which means that must have a length greater or less than . In the diagram below, is the image of under a single transformation of the plane.
Augusta MahapatraWinnifred Pratzler
Rigid Motion and Congruence - MathBitsNotebook(Geo - CCSS Math) Two figures are congruent if and only if there exists one, or more, rigid motions which will map one figure onto the other. (thus maintaining the conditions for the figures to be congruent).
Hasni Posinsk
In a reflection, a mirror image of the figure is created. This does not change angle measure or side length; this is a rigid motion. In a stretch, a figure is enlarged by a scale factor. This changes side length; this is not a rigid motion.
Sanae Angendohr
Explanation: Recall that a rigid motion is that that preserves the distances while undergoing a motion in the plane. Therefore, for the translation to be considered "rigid" the two figures must be congruent by definition of a rigid motion.
Ioana Prentice
There are four types of rigid motions that we will consider: translation , rotation, reflection, and glide reflection. Translation: In a translation, everything is moved by the same amount and in the same direction. Every translation has a direction and a distance.
Merle Vollans
The concepts of congruence, similarity, and symmetry can be understood from the perspective of geometric transformation. Fundamental are the rigid motions: translations, rotations, reflections, and combinations of these, all of which are here assumed to preserve distance and angles (and therefore shapes generally).
Addaia Labbe
A congruence transformation is a transformation that doesn't change the size or shape of an object. There are three main types of congruence transformations, and those are reflections (flips), rotations (turns), and translations (slides).
Milton Tufnall
Exactly equal in size and shape. Congruent sides or segments have the exact same length. Congruent angles have the exact same measure. For any set of congruent geometric figures, corresponding sides, angles, faces, etc. are congruent.
Georgina Lipkin
Translation is a term used in geometry to describe a function that moves an object a certain distance. The object is not altered in any other way. It is not rotated, reflected or re-sized. In a translation, every point of the object must be moved in the same direction and for the same distance.
Latosha Bahi
If there are no fixed points, then the rigid motion is either a translation (which is not the identity) or a ''glide reflection'' which is a translation (which is not the identity) followed by a reflection about a line parallel to the direction of translation.
Ifeanyi Alisch
A dilation is a transformation that produces an image that is the same shape as the original, but is a different size. A dilation that creates a smaller image is called a reduction. • A description of a dilation includes the scale factor (or ratio) and the center of the dilation.
Otger Zschieschank
Composition is a sort of multiplication operation for the rigid motions of a plane. ∏. We can form the composition of any two rigid motions of ∏ to get a new rigid motion. of ∏. We can compose two rigid motions of the same type: two translations, two.
Onita Algandona
the line y = -x is the point (-y, -x). Reflect over any line: Remember that each point of a reflected image is the same distance from the line of reflection as the corresponding point of the original figure. The line of reflection will lie directly in the middle between the original figure and its image.
Parviz Zappa
A basic rigid transformation is a movement of the shape that does not affect the size of the shape. The shape doesn't shrink or get larger. There are three basic rigid transformations: reflections, rotations, and translations. There is a fourth common transformation called dilation.
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A line is a junction of two points. This means that a line
has a beginning and an end:
The beginning and the end are two distinct points. Based on
this, a line is represented either with two Point values or by four
numbers representing its values on the Cartesian axes. To draw a line, the
Graphics class is equipped with the following overloaded DrawLine()
method:
Public Sub DrawLine(pen As Pen, pt1 As Point, pt2 As Point)
Public Sub DrawLine(pen As Pen, pt1 As PointF, pt2 As PointF)
Public Sub DrawLine(pen As Pen, x1 As Integer, _
y1 As Integer, x2 As Integer, y2 As Integer)
Public Sub DrawLine(pen As Pen, x1 As Single, _
y1 As Single, x2 As Single, y2 As Single)
If the line is represented with natural numbers, its origin
can be specified as a Pointpt1 and its end would be represented
with a Pointpt2. If the line is drawn using floating numbers, you
can start it at one PointFpt1 and end it at another PointFpt2. Otherwise, you can specify the starting point with coordinates (x1,
y1) and the end would be represented with coordinates (x2, y2).
The same type of line can be drawn using decimal values from coordinates (x1,
y1) to coordinates (x2, y2).
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In trapezoid $EFGH,$ $\overline{EF} \parallel \overline{GH},$ and $P$ is the midpoint of side $\overline{EH}$. If the area of triangle $PEF$ is $18$, and the area of triangle $PGH$ is $36$, then find the area of trapezoid $EFGH$.
0 users composing answers..
To find the area of trapezoid EFGH, we can use the fact that the area of a trapezoid is equal to the average of the lengths of the two bases multiplied by the height. Given that P is the midpoint of side EH, we can use the areas of triangles PEF and PGH to find the height of trapezoid EFGH. 1. Since P is the midpoint of EH, we know that the height of trapezoid EFGH is equal to the distance between EF and GH. 2. The area of triangle PEF is given as $18.Let′sdenotethebaseoftrianglePEFasx.Therefore,theheightoftrianglePEFisalsoh,wherehisthedistancebetween\overline{EF}and\overline{GH}.3.Usingtheformulafortheareaofatriangle,wehave\frac{1}{2} \times x \times h = 18.Thus,x \times h = 36.4.Similarly,theareaoftrianglePGHisgivenas36, and let's denote the base of triangle PGH as y. Therefore, y×h=72. 5. Now, the average of the bases of trapezoid EFGH is 21(x+y). Multiplying this by the height h, we get the area of trapezoid EFGH as 21(x+y)×h=21(xh+yh)=21(36+72)=54. Therefore, the area of trapezoid EFGH is $54$ square units.
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$\begingroup$In this recent answer, I determine triangles based on equal median, altitude, and bisector from three vertices. That's a different problem than this one, of course, but the equations $(1)$, $(2)$, $(3)$ give the lengths of those segments in terms of the triangle sides; "all you have to do" is solve the system for the side-lengths in terms of the segment-lengths. (Well, you first need to be sure to adjust the equations so that they all refer to segments emanating from the same vertex.)$\endgroup$
$\begingroup$@JAOFELIX Please avoid writing entire sentences using capital letters. It comes off as quite rude, even if being rude is not your intention. (Some people read it as if the author were shouting.)$\endgroup$
3 Answers
3
Couple of lemmata proven below helps us to concisely elucidate the proposed construction. Unless stated otherwise, we use the expression "$\mathrm{angle\space bisector}$" to denote the $\mathrm{interior\space angle\space bisector}$ of an angle.
$\mathbf{Lemma\space 1.1}$
The altitude and the median dropped from a given vertex of all scalene triangles lie on either side of the interior angle bisector at that vertex.
$\mathbf{Proof\space 1.1}$
Consider $\mathrm{Fig.\space 1}$, where $M, D$, and $H$ are the respective feet of the median, the angle bisector, and the altitude dropped from the vertex $A$ of an scalene triangle $ABC$.
If feet of any two lines mentioned above coincide, then the foot of the remaining line coincides with the feet of the other two.
$\mathbf{Proof\space 1.2}$
For instance, if the foot of the median coincides with that of the angle bisector, we have
$$BD = DC \quad\rightarrow\quad \frac{BC\cdot AB}{AB+CA} = \frac{BC\cdot CA}{AB+CA} \quad\rightarrow\quad AB = CA.$$
This proves that $ABC$ is an isosceles triangle with its apex at $A$. In an isosceles triangle, feet of all three lines mentioned above coincide.
The other cases can be proved using similar arguments.
$\mathbf{Lemma\space 2}$
The point of intersection of the extended angle bisector of a given vertex of a scalene triangle and the perpendicular bisector of the opposite side of that vertex lies on the circumcircle of that triangle.
$\mathbf{Proof\space 2}$
We consider the angle bisector of the $\measuredangle A$ (i.e. $AE$) and the perpendicular bisector of the side $BC$ shown in $\mathrm{Fig.\space 2}$. These two lines meet at $F$. Let $\measuredangle BCA = \phi$ and $\measuredangle CAE = EAB = \alpha$. Then $\measuredangle CEF$, which is one of the exterior angles of the triangle $AEC$ is equal to $\left( \alpha + \phi\right)$. This is also one of the exterior angles of the triangle $DFE$. Therefore,
$$\measuredangle DFE = \alpha + \phi – 90^o. \tag{3}$$
Let $O$ be the circumcenter of the triangle $ABC$. Hence, the perpendicular bisector of the side $BC$ (i.e. $DF$) passes through $O$.
We can write that $\measuredangle BOA$, the angle subtended at $O$ by the side $AB$, is equal to $2\phi$. Since $OA = OB$, $OAB$ is an isosceles triangle. Therefore, $\measuredangle OAB$ is equal to $90^o - \phi$, which means that
$$\measuredangle EAO = \alpha + \phi – 90^o. \tag{4}$$
Equations (3) and (4) confirm that $OFA$ is an isosceles triangle. Therefore, $OF = OA$ = Circum-Radius - meaning $F$ lies on the circumcircle of $ABC$.
Please note that this lemma is not applicable to isosceles and equilateral triangles, because it is not possible to define the point $F$.
$\mathbf{Construction}$
The construction of the triangle $ABC$ is carried out in two separate stages. In the first stage, the line, on which the side $BC$ lies, is found after line segments representing the given altitude, angle bisector, and median are laid out in space. In the second stage, the circumcircle of $ABC$ is constructed after finding its center and a point that lies on its circumference. The two vertices $B$ and $C$ are the points of intersection between the circumcircle and the line that contains the side $BC$.
$\mathbf{Stage\space 1}$
We make use of the fact that side $BC$, altitude, and angle bisector forms a right triangle to lay out these three lines in space as shown in $\mathrm{Fig.\space 3}$. First, a circle having $AD$ as its diameter is drawn with its center at $P$, which is the midpoint of the angle bisector $AD$. A second circle is drawn having the length of the altitude as its radius and $A$ as its center. Any one of the two points of intersection between these two circles can be selected as $H$, the foot of the altitude. The line $HD$ contains the side $BC$.
Now, construct another circle having the length of the median as its radius and $A$ as the center to cut the extended $HD$ at $M$ and $N$. In accordance with Lemma 1.1, we have to select $AM$ as the median. If we select $AN$ instead, we are putting altitude and median on the same side of angle bisector. Selection of $AM$ as the median define $M$ as the midpoint of side $BC$.
$\mathbf{Stage\space 2}$
Draw the perpendicular line $MF$ to $HD$ at $M$ to intersect the extended angle bisector $AD$ at $F$ as depicted in $\mathrm{Fig.\space 4}$. According to Lemma 2, $F$ is located on the circumcircle of the sought triangle $ABC$. Therefore, $AF$ is a chord of this circumcircle, the center of which lies on $EQ$, the perpendicular bisector of $AF$. Furthermore, since $M$ is the midpoint of side $BC$ and $MF$ is perpendicular to the side $BC$, the circumcenter of $ABC$ lies on $MF$ as well. This means that the point of intersection of $EQ$ and $MF$ is the circumcenter $O$ of $ABC$. Now, to complete the construction, draw the circumcircle, which has the length of $AO$ as its radius and $O$ as its center to cut the extended $HD$ at $B$ and $C$.
The above described construction produces a unique triangle, if an only if $m \gt d \gt h \gt 0$. The case mentioned in Lemma 1.2, i.e. $m = d = h \gt 0$, where the sought triangle is either an isosceles or an equilateral triangle, can lead to infinite number of solutions. Collapsing of altitude, median, and angle bisector on to a single line makes this case an underdetermined problem and allows the side $BC$ to have any value.
Stage 1 of the construction could have been carried out in two more ways. Firstly, instead of the right triangle already mentioned, we could have constructed the right triangle formed by side $BC$, altitude, and median and continued accordingly. Secondly, since both right triangles have altitude as one of their sides, it is also possible to copy one of them on to the other while observing Lemma 1.1. The last method has an advantage over the other two because we do not have anything to exclude.
At the end of the stage 1 of our construction, we have excluded the median $AN$ (see $\mathrm{Fig.\space 3}$) from our solution space citing a violation of Lemma 1.1. Nevertheless, one can carry out the stage 2 of the construction taking $AN$ as the median to obtain a triangle as the solution, if $h$, $m$, and $d$ satisfies the following condition.
$$\frac{1}{h^2} \ge \frac{1}{m^2} + \frac{1}{d^2} \tag{5}$$
This triangle turns out to have the same altitude and median as the sought triangle. But, the prescribed length of the angle bisector corresponds to that of the exterior angle bisector. This outcome is possible and correct because Lemma 1.1 is not applicable to the bundle of altitude, median, and exterior angle bisector. If the values of $h$, $m$, and $d$ upholds the equal sign of (5), (e.g. $h=12$, $m=20$, and $d=15$), the resulting triangle is the degenerated triangle with $BC=0$.
Excluding $b,c$ from \eqref{1}-\eqref{3},
we get a quadratic expression in $a^2$
\begin{align}
(a^2)^2-8\,(m_a^2-2h_a^2)\,a^2
&+\frac{16(m_a^2-\beta_a^2)(\beta_a^2\,m_a^2-\beta_a^2\,h_a^2-m_a^2\,h_a^2)}{\beta_a^2-h_a^2}
=0
\tag{4}\label{4}
,
\end{align}
which gives the value of the side length $a$.
Equation \eqref{1} provides the value of $b^2+c^2$ in terms of $a$,
and \eqref{3} provides the value of $b^2c^2$ in terms of $a$,
which gives another quadratic equation with the roots $b^2,c^2$.
So the expressions for
the side length $a$ and
the other two side lengths
in terms of $a$ are
$\begingroup$@gcov It is good that you did some follow-up to your own answer. When I drew the triangle using $a_2, b_2,$, and $c_2$, I found that the angle bisector is longer than 4. However, Aretino's answer to your question does not shed any light on why you get this second real solution in the first place. Using the eaqution (5) of your answer I found that, If $$\frac{1}{h^2}\ge\frac{1}{m^2}+\frac{1}{\beta^2},$$ there is always two real solutions. Do you have any idea why? The second reason why I am writing this comment is given in my next comment. $$\mathrm{Contd.}$$$\endgroup$
$\begingroup$@gcov I have already found a geometric construction, which uses straight edge and compasses, to solve this problem. I also get 2 different solutions if the mentioned condition is true. I cannot find a reason to reject this $2^{nd}$ solution without first drawing the corresponding angle bisector. I would like to show this to you, but, unfortunately, we have been barred from answering this question. Is there anything you (you have more privileges than I do) can do to re-open the question to receive answers? Please read my yesterday's comment and a reply to it by Blue.$\endgroup$
$\begingroup$@YNK: I've already voted to reopen the question, there are 4 votes at the moment, just one more needed. Need to look closely at your comment on $a_2$. Can you compose your these comments into the answer to the follow-up question?$\endgroup$
In the indicated case it is given that the median $m_a$ from $A$ and the angle bisector $w_A$ at $A$coincide. In this case the triangle is isosceles, and the altitude from $A$coincides with the line $m_a=w_A$. It follows that the three occurring lengths coincide, and are equal to the height $h_a=|AM_a|$, where $M_a$ is the midpoint of $BC$ on the base of the triangle. When the given lengths $m_a$, $\beta_a$, and $h_a$ do not coincide there is no triangle having the desired properties.
On the other hand, when $m_a=\beta_a=h_a$, the given information does not determine the two points $B$, $C$ at equal distance from $M_a$ on the base.
$\begingroup$@Chritian Blater thanks for the effort you have put but in the question we are never given that the length of angle bisector, median and altitude is equal. We have to find something in general.$\endgroup$
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1. A complement of angle X measures 370. What is the measure of a supplement of angle X?
A) 130 0 B) 143 0 C) 127 0 D) 37 0 E) 53 0
2. Parallelogram DEFG has a base of 12 centimeters and a height of 5 centimeters. What is the area of parallelogram DEFG?
A) 60cm3 B) 60cm2 C) 34cm2 D) 7cm2 E) 120cm2
3. If WXYZ is a parallelogram, which of the following statements is never true.
A) The diagonals of WXYZ bisect each other. B) The consecutive angles of WXYZ are complementary. C) Pairs of opposite sides are parallel. D) Pairs of opposite sides are congruent.
4. A right triangle has legs 5 meters and 12 meters long. What is the length of the hypotenuse?
A) 12m B) 16m C) 14m D) 15m E) 13m
5. To the nearest tenth of a centimeter, what is the circumference of a circle with a diameter of 15 centimeters?
A) 22.1cm B) 47.1cm C) 23.6cm D) 71.7cm E) 94.1cm
6. Two sides of a triangle measure 3 feet and 6 feet. Which of the following could not represent the length of the third side?
A) 8ft B) 3ft C) 7ft D) 9ft E) 4ft
7. Lines m and n lie in the same plane. The slope of line m is -5/3 and the slope of line n is 3/5. How are lines m and n related?
A) Lines m and n are perpendicular. B) Lines m and n are parallel. C) Lines m and n intersect but are not perpendicular. D) Lines m and n coincide.
8. One vertex of a square whose sides are 9 units long has a vertex at (3, 3). Which of the points below could be another vertex of the square?
A) (3, 6) B) (3, 3) C) (3,-6) D) (9,0) E) (-3,3)
9. The lengths of the sides of triangle ABC are 4, 5, and 6. Which set of numbers gives the lengths of the sides of a triangle similar to triangle ABC?
A) (2,5,6) B) (6,7,8) C) (7,8,10) D) (9,10,12) E) (10,20,31)
10. You receive two standard postcards from your friend while she is on summer vacation. The first postcard is 5 inches long and 3.5 inches high. The second postcard is 6 inches long and 4.25 inches high. Which of the following statements is true?
A) The postcards would be similar rectangles if the first postcard were 4.9 inches long. B) The postcards are congruent rectangles. C) The postcards are similar rectangles D) The postcards are congruent squares. E) The postcards would be similar if the height of the second postcard were 4.2 inches.
11. If a tangent and a chord intersect at a point on a circle, what do you know must be true about the measures of the two angles that are formed?
A) They are supplementary. B) They each measure one half the difference of the measures of their intercepted arcs. C) Their measures equal the measures of their intercepted arcs. D) Their measures are one half the measures of their vertical angles.
12. The measure of each interior angle of a regular polygon is 135 0. How many sides does the polygon have?
A) 4 B) 15 C) 8 D) 12 E) 9
13. The measure of each interior angle of a regular polygon is 120 0. How many sides does the polygon have?
A) 7 B) 8 C) 4 D) 6 E) 9
14. If two angles are complementary, then the sum of their measures is 90 0.
A) False B) True
15. The measure of Supplementary angles is 180 0.
A) False B) True
16. The right triangles have exactly one angle that its measure is 90 0.
A) False B) True
17. An Isosceles triangle has exactly two angles that their measures add to 90 0.
A) True B) False
18. The parallel lines never intersect.
A) False B) True
19. Perpendicular lines form angles that measure 180 0.
A) True B) False
20. Point, line, and plane are the three undefined terms in Geometry.
A) True B) False
21. A _________ is the most basic building block of geometry.
A) Circle B) Segment C) Kite D) Point E) Square
22. A __________ is a straight, continuous arrangement of infinitely many points.
A) Segment B) Line C) Sphere D) Chord E) Ray
23. A _________ has length and width, but no thickness.
A) Line B) Square C) Plane D) Rectangle
24. _________ means on the same line.
A) Coplanar B) Collinear C) Segment D) Line
25. ___________ means on the same plane.
A) Sphere B) Plane C) Square D) Coplanar E) Rectangle
26. A ________ begins at a point and extends infinitely in one direction.
A) Plane B) Ray C) Segment D) Diagonal E) Line
27. An _________ is formed by two rays that share a common endpoint.
A) Ray B) Angle C) Diagonal D) Line E) Segment
28. A design for a garden has the shape of a regular octagon. The design is 32 centimeters on a side and apothem 39 cm. What are the perimeter and the area of the design? -Solve the problem and explain in complete sentence the way that you found the solution.
29. A square has an area of one hundred square centimeters. What is the length of each of its sides?
A = 4a
100 = 4a
100/4 = 4a/4
25 = a
The length of each of the side is 25.
-Alex solve this problem, but when he show all work to his teacher, she said that he have a mistake. Write in complete sentence to try to explain Alex where he make a mistake.
30. Now is your time to show you creativity.
Part 1. Create you own word problem about one thing that you was learned in Geometry.
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THE CENTROID OF A TRIANGLE
Suppose that ∆ABC is any triangle. To determine its centroid, draw the median passing through one of its vertices, A. Let D be the midpoint of the line segment BC. (So the median passes through A and D.) Similarly, draw the median passing through C and let the median meet AB at E, the midpoint of the line segment AB. AD intersects CE at T. Then T is the centroid of ∆ABC.
If the length of line segment AT is AT and the length of line segment TD is TD, what is AT:TD? Also, if the length of line segment CT is CT and the length of line segment TE is TE, what is CT:TE?
To answer this, let:
Note that:
Accordingly, we have the following:
……………………………. (*)
As , the equation (*) is equivalent to:
…………………………………….. (**)
The equation (**) implies and .
Hence AT : TD = CT : TE = 2 : 1.
The next question:
If the median passing through B is drawn, does it pass through T? In other words, are all the medians concurrent? Yes! Let's prove it. Let the midpoint of the line segment AC is P. (See the figure below.)
Note that:
The fact that shows that: 1) the points B, T, and P are collinear, and 2) BT : BP = 2 : 1. To sum up, all the medians of ∆ABC are concurrent. All the medians meet at one and only one common point T.
The next question:
Given the coordinates of A, B, and C of the triangle ∆ABC on a cartesian space, how to determine the coordinate of the triangle's centroid? The previous result shows that . The equation implies:
where , , and are the position vectors of T, E, and C, respectively. Since E is the midpoint of the line segment AB, it follows that:
Accordingly, the coordinate of the centroid can be determined easily from the position vector .
Example 1
Given ΔABC with A(1,5), B(-3,8), C(5,11). Find the coordinates of its centroid.
Answer
Let the centroid be T with the position vector .
Consequently, the coordinates of the centroid is (1,8).
Example 2
Given ΔABC with A(1,5,0), B(-3,8,4), C(5,11,2). Find the coordinates of its centroid.
Answer
Let the centroid be T with the position vector .
Consequently, the coordinates of the centroid is (1,8,2).
The next question:
What is the distance between the centroid of a triangle and the base? Consider the ΔABC below.
In the ΔABC, AD and CE are medians and T is the centroid. Our previous result states that CT : TE = 2 : 1. Let K be the point on the line segment AB such that CK ⊥ AB. Let L be the point on the line segment AB such that TL ⊥ AB. Note that ΔCKE is similar to ΔTLE, and consequently:
Since CT : TE = 2 : 1, CE = 3 TE and furthermore . This leads to the conclusion that the distance between the centroid of a triangle and the base of the triangle is one third of the length of the altitude drawn from the vertex opposite to the base.
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interior and exterior angles of a triangle worksheet kuta
Exterior And Interior Angles Of A Triangle Worksheet Answers – Triangles are among the most fundamental shapes in geometry. Knowing how triangles work is essential to understanding more advanced geometric concepts. In this blog post it will explain the different kinds of triangles Triangle angles, how to calculate the perimeter and area of a triangle, and present an example of every. Types of Triangles There are three types to triangles: the equilateral, isosceles, and scalene. Equilateral … Read more
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Class 12 Maths NCERT Solutions for Chapter 11 Three Dimensional Geometry includes solutions to all Miscellaneous Exercise problems. Three Dimensional Geometry Class 12 NCERT Solutions Miscellaneous Exercises are based on the ideas presented in Maths Chapter 11. This activity is crucial for both the CBSE Board examinations and competitive tests. To perform well on the board exam, download the latest CBSE Class 12 Maths Syllabus in PDF format, which are solved by experts to help you understand easily.
2. Find the equation of a line parallel to x-axis line passing through the origin.
Ans: As it is given that the line is passing through the origin and is also parallel to x-axis is x-axis,
Now,
Let us consider a point on x-axis be $\text{A}$
So, the coordinates of $A$ will be $\left( \text{a,0,0} \right)$
Now, the direction ratios of $\text{OA}$ will be,
$\Rightarrow \left( \text{a-0} \right)\text{=a,0,0}$
The equation of $\text{OA}$$\Rightarrow \frac{\text{x-0}}{\text{a}}\text{=}\frac{\text{y-0}}{\text{0}}\text{=}\frac{\text{z-0}}{\text{0}}\Rightarrow \frac{\text{x}}{\text{1}}\text{=}\frac{\text{y}}{\text{0}}\text{=}\frac{\text{z}}{\text{0}}\text{=a}$
Therefore, the equation of the line passing through origin and parallel to x-axis is $\frac{\text{x}}{\text{1}}\text{=}\frac{\text{y}}{\text{0}}\text{=}\frac{\text{z}}{\text{0}}$.
3.if the lines $\frac{\text{x-1}}{\text{-3}}\text{=}\frac{\text{y-2}}{\text{2k}}\text{=}\frac{\text{z-3}}{\text{2}}$ and $\frac{\text{x-1}}{\text{3k}}\text{=}\frac{\text{y-1}}{\text{1}}\text{=}\frac{\text{z-6}}{\text{-5}}$ are perpendicular Find the value of k
Ans: From the given equation we can say that ${{\text{a}}_{\text{1}}}\text{=-3,}{{\text{b}}_{\text{1}}}\text{=2k,}{{\text{c}}_{\text{1}}}\text{=2}$and ${{\text{a}}_{\text{2}}}\text{=3k,}{{\text{b}}_{\text{2}}}\text{=1,}{{\text{c}}_{\text{2}}}\text{=-5}$.
We know that the two lines are perpendicular, if ${{\text{a}}_{\text{1}}}{{\text{a}}_{\text{2}}}\text{+}{{\text{b}}_{\text{1}}}{{\text{b}}_{\text{2}}}\text{+}{{\text{c}}_{\text{1}}}{{\text{c}}_{\text{2}}}\text{=0}$
Therefore, the shortest distance between the above two lines is of $\text{9}$ units.
5. Find the vector equation of the line passing through the points $\left( \text{1,2,-4} \right)$ and perpendicular to the two lines $\frac{\text{x-8}}{\text{3}}\text{=}\frac{\text{y+19}}{\text{-16}}\text{=}\frac{\text{z-10}}{\text{7}}$ and $\frac{\text{x-15}}{\text{3}}\text{=}\frac{\text{y-29}}{\text{8}}\text{=}\frac{\text{z-5}}{\text{-5}}$
Ans: According to the question, we get that ${\vec{b}=}{{\text{b}}_{\text{1}}}{\hat{i}+}{{\text{b}}_{\text{2}}}{\hat{j}+}{{\text{b}}_{\text{3}}}{\hat{k}}$ and ${\vec{a}=\hat{i}+2\hat{j}-4\hat{k}}$
We know that the equation of the line passing through point and also parallel to vector, we get
Conclusion
NCERT solutions for class 12 maths Three Dimensional Geometry provide some examples of problems covered in the miscellaneous exercise?
The miscellaneous exercise covers a wide range of topics in 3D geometry, including:
Direction cosines and ratios of lines
Finding the equation of a line (Cartesian and vector forms)
Identifying co-planar and skew lines
Calculating the angle between two lines
Finding the shortest distance between two lines (parallel and skew)
Equation of a plane (normal form, intercept form, etc.)
Determining the angle between two planes
Finding the distance between a point and a plane
2 11 Miscellaneous Exercise 3D Geometry."
3. Three Dimensional Geometry sample questions?
Find the direction cosines of the line joining the points A(2, 1, 3) and B(4, -1, 2).
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Answer
Yes, they are right angles
Work Step by Step
The legs are 50 and 120 and the hypotenuse is 130. Plug them into the Pythagorean theorem to determine if they are equal $a^{2}$ + $b^{2}$= $c^{2}$-->$50^{2}$ + $120^{2}$= $130^{2}$:
$50^{2}$ + $120^{2}$= $130^{2}$ -simplify the exponents-
2500+14400=16900 -add like terms-
16900=16900 -Compare,they are equal so this is a right triangle-
Yes, they are right angles.
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When can we say that an angle is positive or negative?
Answered by Michael Wilson
An angle can be considered positive or negative depending on the direction in which it rotates. When measuring angles, we typically use a reference point called the initial side and a reference direction called the positive x-axis.
If an angle rotates in a counterclockwise or anticlockwise direction, it is considered positive. This means that as the angle increases, it moves in the direction of the positive x-axis. For example, if we start at the positive x-axis and rotate 45 degrees counterclockwise, the resulting angle would be positive 45 degrees.
On the other hand, if an angle rotates in a clockwise direction, it is considered negative. This means that as the angle increases, it moves in the opposite direction of the positive x-axis. For instance, if we start at the positive x-axis and rotate 30 degrees clockwise, the resulting angle would be negative 30 degrees.
To understand this concept further, let's consider some real-life examples. Imagine you are standing at a crossroad, facing north. If you turn to your left and face west, you have rotated in a counterclockwise direction, and the angle you have formed is positive. Conversely, if you turn to your right and face east, you have rotated in a clockwise direction, and the angle you have formed is negative.
In mathematics, we can represent angles using different units of measurement. The most common units are degrees and radians. In the degree method, a full rotation is 360 degrees, with positive angles measured counterclockwise and negative angles measured clockwise.
In the radian method, a full rotation is equal to 2π radians, where π is approximately 3.14159. Similar to degrees, positive angles in radians are measured counterclockwise, and negative angles are measured clockwise.
It's important to note that angles can also be measured beyond a full rotation. These angles are called coterminal angles. Coterminal angles share the same initial and terminal sides but differ in the number of complete rotations they make. For example, an angle of 390 degrees and an angle of 30 degrees are both coterminal angles because they end at the same position.
To summarize, an angle is considered positive when it rotates counterclockwise from the initial side, and it is negative when it rotates clockwise. The choice of units (degrees or radians) determines the specific numerical value assigned to the angle. Understanding the concept of positive and negative angles is crucial in various fields, including geometry, physics, and engineering
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Right
Interior
Exterior
Adjacent
Vertical
Complementary
Supplementary
Dihedral
A dihedral angle is the angle between two intersecting planes or half-planes. In chemistry, it is the clockwise angle between half-planes through two sets of three atoms, having two atoms in common. In solid geometry, it is defined as the union of a line and two half-planes that have this line as a common edge. In higher dimensions, a dihedral angle represents the angle between two hyperplanes.
The planes of a flyi...
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Coordinate Geometry Essentials: Unveiling Cartesian Coordinates
January 13, 2024
Tina Howes
United States of America
Geometry
Tina Howes, a Stanford University mathematics graduate, brings a decade of expertise to math assignment help. Her passion for precision and depth in mathematical concepts ensures students receive unparalleled assistance. With a commitment to excellence.
Coordinate geometry, also referred to as analytic geometry, stands as a pivotal branch of mathematics seamlessly intertwining algebraic techniques with geometric intuition. At its core, this mathematical discipline finds its foundation in the Cartesian coordinate system, a revolutionary tool introduced by René Descartes in 1637, profoundly altering the landscape of mathematical thinking. This blog embarks on a comprehensive exploration of the essentials within coordinate geometry, peeling back the layers to unravel the intricacies inherent in Cartesian coordinates. These coordinates, expressed as (x, y), serve as the key to representing points in a two-dimensional plane, where the x-axis and y-axis intersect at the origin (0, 0). As we navigate through the fundamental principles of plotting points and graphing equations on the Cartesian plane, we uncover a visual language that transforms abstract mathematical concepts into tangible geometric shapes. Essential to this understanding are the distance formula and midpoint formula, crucial tools employed in various applications, ranging from physics to engineering. The blog further delves into the equations of lines, exploring the slope-intercept form and point-slope form, enabling a nuanced comprehension of linear relationships. The identification of parallel and perpendicular lines, based on their slopes, adds another layer of analytical prowess. Moving beyond lines, the exploration extends to conic sections—the circle, ellipse, and hyperbola—each with its unique Cartesian equation, offering insights into their properties and applications across disciplines.
As the narrative unfolds, transformations in the coordinate plane, including translation, rotation, reflection, and dilation, emerge as powerful tools to analyze geometric figures' dynamic changes. These transformations, governed by mathematical principles, find applications in fields as diverse as computer science, where Cartesian coordinates are fundamental to graphics rendering, and economics, where they visualize and model economic relationships. The profound impact of coordinate geometry extends beyond theoretical realms, permeating diverse sectors such as physics, engineering, computer science, and economics. This dynamic interplay between algebraic precision and geometric visualization underscores the versatility and significance of Cartesian coordinates. In essence, the Cartesian coordinate system transcends its historical origins, becoming an indispensable ally in the quest for mathematical understanding and problem-solving prowess, shaping the very fabric of how we perceive and navigate the mathematical landscape. If you're looking to solve your Geometry assignment, understanding the applications and importance of Cartesian coordinates can provide valuable insights and enhance your problem-solving skills in this mathematical discipline.
Historical Context
In the historical context of coordinate geometry, the 17th-century emergence of the Cartesian coordinate system marked a transformative juncture in mathematical thought. René Descartes, a polymathic French philosopher and mathematician, laid the foundation for this revolutionary approach in his 1637 work "La Géométrie." Descartes sought to bridge the realms of algebra and geometry, introducing the concept of representing geometric figures algebraically. The Cartesian coordinate system, consisting of perpendicular x and y axes intersecting at the origin (0, 0), became the cornerstone of this integration. Descartes's innovation not only revolutionized mathematical thinking but also laid the groundwork for a unified analytical language applicable across various disciplines. The coordinates (x, y) assigned to each point in the plane allowed for a systematic representation of geometric shapes, unleashing a new era in mathematical exploration and problem-solving. The historical roots of the Cartesian coordinate system highlight its enduring impact on the evolution of mathematical methodologies and its pervasive influence on fields ranging from physics to computer science.
Descartes's Visionary Integration:
Descartes's visionary integration of algebra and geometry in the Cartesian coordinate system was a groundbreaking endeavor that aimed to create a seamless connection between numerical calculations and visual representations. By assigning numerical coordinates to points in the plane, Descartes effectively transformed geometric problems into algebraic equations, fostering a new way of approaching mathematical challenges.
Transformation of Mathematical Language:
The introduction of the Cartesian coordinate system represented a transformative shift in mathematical language and methodology. Prior to Descartes, geometry relied heavily on geometric constructions and theorems, whereas algebra dealt with abstract numerical relationships. The Cartesian coordinates provided a unifying language, enabling mathematicians to express geometric concepts through equations and facilitating a more systematic and interconnected approach to mathematical analysis. This transformation not only streamlined mathematical communication but also paved the way for the development of analytical geometry and its wide-ranging applications in diverse scientific and engineering disciplines.
The Cartesian Coordinate System
The Cartesian Coordinate System, a cornerstone of coordinate geometry, establishes a two-dimensional framework that intertwines algebraic principles with geometric concepts. Envisioned by René Descartes in the 17th century, this system introduces a pair of perpendicular axes, the x-axis and the y-axis, intersecting at the origin (0, 0). Through these axes, every point on the plane is uniquely defined by coordinates (x, y), symbolizing horizontal and vertical distances from the origin, respectively. The act of plotting points and graphing equations becomes a visual manifestation of mathematical relationships on the Cartesian plane. The Distance Formula, \(\sqrt{(x₂ - x₁)^2 + (y₂ - y₁)^2}\), and the Midpoint Formula, \(\left(\frac{x₁ + x₂}{2}, \frac{y₁ + y₂}{2}\right)\), emerge as fundamental tools, facilitating calculations crucial in various mathematical applications. Equations of lines, such as the Slope-Intercept and Point-Slope forms, empower mathematicians to articulate linear relationships elegantly. As we unravel the complexities of the Cartesian Coordinate System, its historical significance and practical utility unfold, shaping mathematical landscapes and fostering a profound connection between algebra and geometry.
Understanding the Basics
The Cartesian coordinate system is a two-dimensional plane formed by two perpendicular axes: the x-axis and the y-axis. The point of intersection, known as the origin, is denoted as (0, 0). Each point in the plane is uniquely identified by its coordinates (x, y), where x represents the horizontal distance from the origin along the x-axis, and y represents the vertical distance along the y-axis.
Plotting Points and Graphing Equations
To graphically represent equations and geometric shapes, we use the Cartesian plane. Plotting points involves moving along the x-axis and then the y-axis based on the given coordinates. Graphing equations involves identifying patterns and plotting multiple points to sketch curves, lines, or other shapes.
Distance Formula and Midpoint Formula
Two fundamental formulas in coordinate geometry are the distance formula and the midpoint formula. The distance formula calculates the distance between two points (x₁, y₁) and (x₂, y₂) using the formula:
[ d = \sqrt{(x₂ - x₁)^2 + (y₂ - y₁)^2} ]
The midpoint formula finds the midpoint between two points:
[ \left(\frac{x₁ + x₂}{2}, \frac{y₁ + y₂}{2}\right) ]
These formulas are essential tools in various mathematical applications, such as physics and engineering.
Equations of Lines
In the realm of coordinate geometry, the study of equations of lines constitutes a pivotal chapter. Linear equations find expression in various forms, with the slope-intercept form \(y = mx + b\) standing as a cornerstone. Here, \(m\) denotes the slope, signifying the rate of change, and \(b\) represents the y-intercept, the point where the line intersects the y-axis. Understanding this form not only facilitates the graphical representation of lines but also provides immediate insights into their behavior. Another essential representation is the point-slope form \(y - y₁ = m(x - x₁)\), which emphasizes a specific point \((x₁, y₁)\) on the line and its corresponding slope \(m\). This form proves particularly useful when the slope and a point on the line are known. Beyond these foundational forms, coordinate geometry offers a seamless method for discerning parallel and perpendicular lines through slope analysis, unraveling the geometric intricacies encoded within the equations, and paving the way for advanced mathematical applications.
Slope-Intercept Form
One of the most common linear equation forms is the slope-intercept form, y=mx+b. In this equation, m represents the slope, indicating the steepness of the line, and b is the y-intercept, the point where the line crosses the y-axis. This form facilitates a quick interpretation of a line's characteristics. By understanding the slope and y-intercept, one gains insights into how the line behaves and intersects with the coordinate axes.
Point-Slope Form
The point-slope form of a line is 1)y−y1=m(x−x1), where (1)(x1,y1) is a specific point on the line, and m is the slope. This form is particularly useful when the slope and a point on the line are known. It provides a straightforward way to express linear equations, emphasizing the relationship between a point on the line and its slope.
Parallel and Perpendicular Lines
Coordinate geometry offers a clear method for identifying parallel and perpendicular lines. Two lines are parallel if they have the same slope, reflecting similar steepness. In contrast, perpendicular lines have slopes that are negative reciprocals, resulting in a 90-degree angle between them. This understanding is crucial in various applications, from designing structures to solving real-world problems where the relationship between lines is a key factor in the analysis.
Conic Sections
In the realm of coordinate geometry, conic sections take center stage, captivating mathematicians and scientists alike with their elegant curves and profound applications. A conic section is the intersection of a plane with a cone, resulting in distinct shapes—circles, ellipses, hyperbolas, and parabolas. The circle, a symmetrical figure with points equidistant from its center, possesses a unique simplicity encoded in its Cartesian equation \((x - h)² + (y - k)² = r²\). Ellipses, characterized by their elongated or compressed forms, reveal their secrets through the equation \(\frac{x²}{a²} + \frac{y²}{b²} = 1\), where \(a\) and \(b\) denote the semi-major and semi-minor axes. Hyperbolas, with their dynamic, open curves, boast equations \(\frac{x²}{a²} - \frac{y²}{b²} = 1\) or \(\frac{y²}{b²} - \frac{x²}{a²} = 1\). Finally, parabolas, the curve of countless applications, emerge through equations of the form \(y = ax² + bx + c\), each conic section offering a mathematical lens through which the intricacies of space, physics, and beyond can be explored and understood.
The Circle
In coordinate geometry, the circle is a fundamental geometric shape represented by the equation (x−h)2+(y−k)2=r2, where (ℎ,)(h,k) is the center of the circle, and r is the radius. This equation provides a concise means of describing the locus of points equidistant from the center. Circles are prevalent in mathematical analysis, physics, and engineering, with applications ranging from planetary orbits to the design of circular structures. Understanding the circle's equation and properties is crucial for solving problems involving curvature, distance, and symmetry.
The Ellipse
The ellipse, another conic section, is expressed in the Cartesian coordinate system by the equation 2=1a2x2+b2y2=1, where a and b represent the semi-major and semi-minor axes, respectively. This equation highlights the geometric features of an ellipse, emphasizing its elongation along the major axis. Ellipses find applications in astronomy, physics, and engineering, playing a significant role in describing the orbits of celestial bodies and designing optical instruments. Analyzing and manipulating ellipses in coordinate geometry enhances our understanding of their diverse applications across different scientific disciplines.
The Hyperbola
The hyperbola, a conic section with distinctive branches, is represented in standard form by 2=1a2x2−b2y2=1 or 2=1b2y2−a2x2=1, depending on its orientation. Hyperbolas exhibit unique properties, with applications in physics and engineering, particularly in the study of electromagnetic waves and satellite trajectories. Understanding the equation and characteristics of hyperbolas is essential for analyzing their asymptotic behavior and predicting the paths of particles in a variety of physical systems. The study of hyperbolas in coordinate geometry enriches our ability to model and comprehend complex phenomena in the natural and applied sciences.
Transformations in the Coordinate Plane
Transformations in the coordinate plane constitute a pivotal aspect of coordinate geometry, offering a profound understanding of how geometric shapes evolve and interact. These transformations include translations, rotations, reflections, and dilations, each playing a unique role in reshaping figures. A translation involves shifting a figure horizontally or vertically by adding or subtracting constant values from the coordinates of its points. Rotations, on the other hand, entail turning a figure about the origin or a specific point, necessitating a grasp of trigonometry and the unit circle for a comprehensive comprehension of rotational transformations. Reflections flip a figure over an axis of reflection, altering the coordinates based on the chosen axis. Dilation, the resizing of a figure while maintaining its shape, relies on a scale factor to determine the degree of enlargement or reduction. These transformations not only contribute to the aesthetic aspects of geometry but also find practical applications in fields such as physics, engineering, and computer science, where understanding how objects change and relate in space is crucial for analysis and problem-solving.
Translation
Translation, a fundamental transformation in coordinate geometry, involves shifting a figure horizontally or vertically without altering its shape. This process modifies the coordinates of each point in the figure by adding or subtracting constant values. It provides a means to explore spatial changes while preserving the inherent characteristics of the object. Whether applied in graphic design or scientific simulations, translation plays a pivotal role in manipulating and analyzing geometric configurations within the Cartesian coordinate system
Rotation
Rotation, another transformative operation within the coordinate plane, entails turning a figure about the origin or a designated point. This geometric transformation requires a nuanced understanding of trigonometry and the unit circle. As the figure revolves, each point's coordinates are adjusted accordingly, creating a new orientation. This rotational manipulation finds applications in various domains, from engineering, where it aids in designing three-dimensional structures, to physics, where it models the rotation of celestial bodies. The principles of rotation extend beyond geometry, influencing diverse fields such as computer graphics and robotics.
Reflection
Reflection, a symmetry operation in coordinate geometry, involves flipping a figure over a designated line known as the axis of reflection. This transformation alters the coordinates of each point based on the axis of reflection, creating a mirror image of the original shape. Whether applied in art, where symmetry is aesthetically pleasing, or physics, where reflective surfaces impact the behavior of light, understanding the principles of reflection is paramount. This geometric operation plays a pivotal role in diverse fields, influencing not only visual representations but also the analysis of structural and optical properties.
Dilation
Dilation, a transformative process in the Cartesian coordinate system, entails resizing a figure while maintaining its shape. The scale factor determines the degree of enlargement or reduction, allowing for precise control over the transformation. In applications ranging from map scaling to medical imaging, dilation proves invaluable in representing objects at different scales. This geometric manipulation extends its utility beyond mere size adjustments, influencing the study of fractals and mathematical modeling. Dilation provides a versatile tool within coordinate geometry, enabling the exploration of proportional relationships and the analysis of complex spatial structures.
Applications of Coordinate Geometry
Applications of coordinate geometry span across diverse fields, playing a crucial role in solving real-world problems. In physics and engineering, Cartesian coordinates serve as a fundamental tool for analyzing the motion of objects, designing structures, and modeling physical phenomena. In computer science, particularly in graphics and image processing, the precision of Cartesian coordinates is essential for rendering images, defining object positions, and implementing algorithms. Economists utilize coordinate geometry to model economic relationships, employing graphs and equations to visualize concepts such as supply and demand curves and production possibilities. The applications extend beyond these realms, touching areas like biology, where spatial relationships within biological structures are studied, and geography, where coordinates aid in mapping and navigation. From predicting the trajectory of a projectile to optimizing the layout of a city, coordinate geometry provides a versatile framework for problem-solving, making it a cornerstone in the toolkit of professionals across scientific and mathematical disciplines.
Physics and Engineering
Coordinate geometry is foundational in physics and engineering applications, playing a pivotal role in analyzing object motion and describing physical phenomena. Engineers leverage Cartesian coordinates for designing structures, analyzing stress distributions, and modeling complex systems. From tracking the trajectory of projectiles to simulating fluid dynamics, the mathematical precision afforded by coordinate geometry enhances the understanding and prediction of physical processes, making it an indispensable tool in the realm of physics and engineering.
Computer Science
In computer science, Cartesian coordinates are integral to graphic rendering and object positioning. Algorithms driving image processing and computer-aided design heavily rely on coordinate geometry. The precise placement of pixels, the definition of shapes, and the rendering of three-dimensional spaces all hinge on the principles of Cartesian coordinates. From video game design to virtual simulations, the fusion of mathematical precision and computational power makes coordinate geometry a fundamental component in the evolving landscape of computer science and technology.
Economics
In economics, coordinate geometry serves as a valuable tool for modeling and analyzing economic relationships. Graphs and equations, derived from Cartesian coordinates, offer visual representations of supply and demand curves, production possibilities, and other economic concepts. The ability to graphically depict economic data enhances the comprehension of complex relationships, aiding economists in making informed predictions and policy recommendations. Whether visualizing market trends or exploring the impact of policy changes, coordinate geometry provides economists with a powerful means of representing and understanding the intricate dynamics of economic systems.
Conclusion
In conclusion, the Cartesian coordinate system stands as a foundational pillar in the realm of mathematics, seamlessly weaving together algebraic principles and geometric insights. René Descartes' visionary creation in the 17th century, the Cartesian plane, has left an indelible mark on mathematical thinking, providing a universal language for expressing spatial relationships and analytical concepts. From the essential understanding of plotting points and graphing equations to the profound implications of distance and midpoint formulas, coordinate geometry unveils a systematic approach to problem-solving. Equations of lines, conic sections, and transformative operations in the coordinate plane further expand its applications, influencing diverse fields such as physics, engineering, computer science, and economics. The elegant synergy between algebra and geometry, epitomized by Cartesian coordinates, continues to empower mathematicians, scientists, and engineers in unraveling the mysteries of the physical world and solving complex problems. As we reflect on the historical journey and multifaceted applications of coordinate geometry, we appreciate its enduring significance as a unifying force in mathematical exploration and a catalyst for innovation across disciplines.
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Knowledge Check
Any tangent at a point P(x,y) to the ellipse x28+(y2)=1 meets the coordinate axes in the points A and B such that the area of the triangle OAB is least, then the point P is
A(√8,0)
B(0,√18)
C(2,3)
Dnone
Question 2 - Select One or More
If the plane x1+y2+z3=1 meets the co-ordinate axes in the points A, B and C, then area of △ABC in sq. units is
A32
B52
C72
D92
Question 3 - Select One
If a tangent having a slope of −43 to the ellipse x218+y232=1 intersects the major and minor axes in points AandB respectively, then the area of ΔOAB is equal to (A) 12 sq. untis (B) 24 sq. units (C) 48 sq. units (D) 64 sq. units
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Description: HK Math Attack is a mnemonic device used to locate the center of a circle.
Application: Use this acronym to teach students how to find the center of a circle.
Process: Teach students the following formula: (x-H)^2 + (y-K)^2 = R^2. Tell students that the center of the circle is (h,k) and using the formula, students will be able to find the center of any circle and by extension, the radius.
Ex. (x-3)^2 + (y-5)^2 = 100. Tell students to find "h" first (3). Next, direct students to find "k" (5) and using the "h comes first, k comes second" method, students will be able to determine the center of the circle. Remind students to find the radius as well.
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Students will practice finding missing sides in special right triangles with this set of three mazes. This includes 45°-45°-90° and 30°-60°-90° triangles. Each maze includes versions for both simplest radical form and decimal approximations and is slightly more difficult than the previous maze. For simplest radical form, rationalizing the denominator is required. Each maze is slightly more difficult than the previous maze. phenomenal resource for my students to practice what they learned in Special Right Triangles! They enjoyed it!
—KELLY C.
My students were very engaged using this resource following our first introduction to special right triangles. I wanted a resource that would provide immediate feedback for students and this resource was perfect.
—ANITA C.
A fun way for students to practice their special right triangle rules. I love that there are different levels of difficulty. I like to let my students choose what level they do. Excellent resource!
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Vprob1
Problem 1 of 3
A ship travels 15 km on a bearing of 070°. It then changes direction and travels 22 km on a bearing of 200°.
(a) Draw a vector diagram modelling this situation.
(b) How far is the ship from its starting point?
(c) Determine the direction the ship must head in order to return to its starting point. Answer to the nearest degree.
Solution
(a)
(b) To solve this problem we can use the sin law or the cosine law to find the distance of AC(A is the starting point, B is the point where they changes direction and C is the final position of the ship).
Sine Law Solution
a/Sin A = b/Sin B
22 Km/Sin 87.08° = b/Sin 50°
b*Sin 87.08° = 22 km*Sin50°
b = (22 km*Sin50°)/Sin87.08°
b = 16.87 km
Cos Law Solution
b2 = a2+c2-2*a*c*Cos B
b2 = (222+152)-(2*22*15*Cos 50°)
b2 = 709-424.23
b2 = 284.77
b = 16.87 km
The ship is 16.87 km away from the starting point
(c) There are 3 ways to anwers this question, we can use the bearing method, the degrees and compass method or the compass and degrees method.
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Midpoint formula | Analytic geometry (article) | Khan Academy (2024)
Walk through writing a general formula for the midpoint between two points.
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Alexene Faith S. Tomate
8 years agoPosted 8 years ago. Direct link to Alexene Faith S. Tomate's post "Good Day! What if the giv..."
Good Day! What if the given are the other endpoint and the midpoint? How do you get the coordinates of the other endpoint?
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(38 votes)
Stephen
4 years agoPosted 4 years ago. Direct link to Stephen's post "I believe you would simpl..."
I believe you would simply find the differences in x and y from the midpoint to the one endpoint, multiply them by two (giving yourself the two side lengths of a right triangle, if you choose to think about your two points in that way), and add these displacements to your given endpoint.
Comment on Stephen's post "I believe you would simpl..."
(22 votes)
Jade Slowik
8 years agoPosted 8 years ago. Direct link to Jade Slowik's post "How would you solve a pro..."
How would you solve a problem in which you do not know point B but are given the midpoint and point A?
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(15 votes)
John Williamson
8 years agoPosted 8 years ago. Direct link to John Williamson's post "Here's what I did when I ..."
You basically are averaging the X and Y values. Consider if you had a grade of 60 and a grade of 100, how would you find the grade that is halfway between them? You would average them. The Midpoint Formula does the same thing. If one X-value is at 2 and the other X-value is at 8, to find the X-value halfway between them, you add 2+8 and divide by 2 = 5. Your would repeat the process for the Y-values to find the Y-coordinate of the midpoint.
It doesn't tell us how to solve for when you have one point and the midpoint... what do you do for that? Because it only shows solving for the midpoint, but what if you're solving for the other point on the line?
Find the change in Y and change in X between that 2 points that you have. Your point b will be on the opposite side of the midpoint from point a. And it will have the same change in Y and change in X. For example:
If a = (2,5) and the midpoint = (-1,3): Change in Y = 5-3 = 2 Change in X = 2-(-1) = 3 Point a is to the right of the midpoint. So point b must be to the left. So, we move in the opposite direction from the midpoint. To go left, we need to change the values to negatives. y-value of b = 3-2 = 1 x-value of b = -1-3 = -4 b = (-4,1)
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(3 votes)
26jamiras614
a year agoPosted a year ago. Direct link to 26jamiras614's post "I am so confused but okay"
a year agoPosted a year ago. Direct link to Kim Seidel's post "The midpoint formula is b..."
The midpoint formula is basically an average. You add the two x-values and divide by 2. You add the two y-values and divide by 2. This gives you the coordinates of the midpoint (the point located half-way between the original two points).
FAQs
The midpoint formula is just an average. Add the 2 X-values, then divide by 2.Add the 2 Y-values, then divide by 2. You have then found the average for the X and Y values which gives you the point half way between the original 2 pointsThe midpoint theorem states that "The line segment in a triangle joining the midpoint of any two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side."Overview: A midpoint is the exact center point between two defined points. To find this center point, midpoint formula is applied. In 3-dimensional space, the midpoint between (x1, y1, z1) and (x2, y2, z1) is (x1+x2 )/2,(y1+y2 )/2,(z1+z2 )/2.
The midpoint method uses the average or the midpoint between two data points to calculate the percent change in the price of a good and its percent change in quantity supplied or demanded. Those two values are then used to calculate the elasticity of supply and demandStatement: The midpoint theorem states that the line segment joining the midpoints of any two sides of a triangle is parallel to the third side and equal to half of the third side. i.e., in a ΔABC, if D and E are the midpoints of AB and AC respectively, then DE || BC and DE = ½ BC.
The advantage of the midpoint method is that one obtains the same elasticity between two price points whether there is a price increase or decrease. This is because the formula uses the same base for both cases It does not matter which point is point #1 vs. point #2.
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Relative positions of a straight line and a plane
To determine the relative positions of a straight line $$r (A'; \overrightarrow{v})$$ and a plane $$\pi(P;\overrightarrow{u},\overrightarrow{v})$$, we express the straight line by means of its implicit equations and the plane with its general equation:
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