a gardener wants to fence a circular garden of diameter 21 m find the length of the rope he - math - perimeter and area
a gardener wants to fence a circular garden of diameter 21 m
find the length of the rope he needs to purchase, if he makes 2
rounds of fence also find the costs of the rope, if it cost rs 4
per meter (takeπ)
d = 21 m
r =![](https://img-nm%20mnimgs%20com/img/study_content/curr/1/7/3/41/1541/chapter%2011_html_96a2a55%20gif)
circumference = 2πr =![](https://img-nm%20mnimgs%20com/img/study_content/curr/1/7/3/41/1541/chapter%2011_html_521eed1b%20gif)= 66 m
length of rope required for fencing = 2 × 66 m = 132 m
cost of 1 m rope = rs 4
cost of 132 m rope = 4 × 132 = rs 528
the distance to be covered to go one round along perimeter of this quadrilateral is12 cm + 3 cm + 5.6 cm + 19 cm you can find the radius in the following way. 
write the equation of the line that is perpendicular to the given tangent line. this perpendicular goes through the stated center, and intersects the tangent; the perpendicular's slope is the negative reciprocal of the stated tangent line.
find the intersection (point t) of the tangent and radius line--then use the distance formula to find the radius, which is a segment from the stated center to the point t. you don't say that the area must be a rectangle. the true largest area would be a circle, with the circumference equal to 400m and the area would be 12,732.4 m^2. however, evidently, the area is restricted to a rectangle, in which case the maximum area is a square, and the side of the square will be 
400/4 = 100 m. the area is 100x100 = 10,000 m^2.
algebraically,
let x = one pair of opposite sides
(400-2x)/2 = 200-x = the other pair of opposite sides

area = a(x) = x(200 - x) = -x^2 + 200x
x-coordinate of vertex = -b/2a = 200/2 = 100
a(100) = 10,000

that's it! ;) the straight distance between two place on a map can be measured by a strip of paper and divider.use a ruler to measure the distance between the two places. if the line is quite curved, use a string to determine the distance, and then measure the string. find the scale for the map you're going to use. it might be a ruler bar scale or a written scale, in words or numbers. first draw the directions, i recommand you draw them as vectors (like arrows):

- north points to the top
- south points to the bottom
- east points to the right
- west points to the left

now lets draw a line from the center of your "+"-sign at about a 45 degree angle to the top right.

this is the hiker that travels northeast. now the next hiker is simple, you just draw a line from the center to the east. that would be a straight line.

now we assume that the angle between the two hikers is 45 degrees. because that's one piece of the given information.

let's calculate how far the first hiker has traveled: 7,5 x 2 = 15 km.
let's calculate how far the second hiker has traveled: 6 x 2 = 12 km.

now we can use sine to calculate the distance between the two hikers.
note: please set your calculator on degrees. we call hiker's distance: x.

sin(45) = x / 15
15 * sin(45) = x

now we know the distance between the two hikers.

greetx jtk v = (4/3)πr³ 

dv/dt = 4πr².dr/dt

so, when the diameter is 20 cm we have:

-4π = 4πr².dr/dt

hence, dr/dt = -1/r²

i.e. dr/dt = -1/10² => -1/100

now, d = 2r

so, d(d)/dt = 2.dr/dt

=> d(d)/dt = -1/50 cm/hour

or, -0.2 mm/hour

:)> let c1 and c2 are two concentric circles with center o and radius of outer circle is 5 cm.given : ac is a chord with length 8 cm that is tangent to inner circle .to find : radius of inner circle i.e. odac is a tangent for c1 at d so,od ⏊ ac [tangent at a point on the circle is perpendicular to the radius through point of contact ]so, oad is a right-angled triangle at d .also it implies that od is perpendicular to the chord ac in c2so we have,ad = dc [perpendicular from the center to the chord bisects the chord]ac = ad + dc8 = ad + adad = 4 cmin △oad by pythagoras theorem(oa)2 = (od)2 + (ad)2(5)2 = (od)2+ (4)225 = (od)2 + 16(od)2= 9od = 3 cm in this we are asked to divide it in 2:3 ratioso, 3+2 = 55, number congruent of line segment is formed.i. firstly, we have to draw a line segment ab of length 7 cm. draw a line with an acute angle from point aii. point the compass on a, and cut the new line with some radiusplace the compass along the line and make 5 arcs. since we need 5 congruent parts, and name the last one as “g”iii. take length of bg as radius draw an arc from a just below it.again, take the length of ag as radius and draw an arc which cuts the previous arc from b, name the intersection point as hconnect b and hiv. taking the same length that we used in line ag, draw arcs along bh and name the points.v. connect the points along ag and bh. name the intercepts on the line ab as m, n, o & p.vi. thus we can divide a line segment as ao:ob = 6:4= 3:2 given: (i) xy = 4.5cm

(ii) yz = 5 cm

(iii) zx = 6 cm

to construct: a xyz

steps for construction

step 1 : first we will draw a line segment xy of length 4.5cm



step 2 : we have to draw an arc , taking x as center and radius 6cm



step 3 : we have to another arc, with radius 5cm and center at y and mark the intersection of arcs as z.



step 4 : we have to join xz and xy



this is the required triangle consider the following figure:two circles with centres q and r touch each other externally at x. another circle, with centre p, touches the other two circles internally at s and t. if pq = 8 cm, qr = 11 cm and pr = 9 cm, find the radius of circle r.