if the sides of a triangle are 6 cm, 8 cm and 10 cm, respectively, then determine whether the triangle is a right angle triangle or not.
the sides of the triangle are 6 cm, 8 cm and 10 cm. the longest side is 10 cm. (10)2 = 100 ….(i) now, the sum of the squares of the other two sides will be, (6)2 + (8)2 = 36 + 64 = 100 ….(ii) (10)2 = (6)2 + (8)2 ….. from (i) and (ii) by the converse of pythagoras theorem: the given sides form a right angled triangle.
hello, red.since the triangle is a right triangle, you can use the pythagorean theorem.let x be the length of the unknown side.then,92 + x2 = (9√2)281 + x2 = 81·2 = 162x2 = 162 - 81x2 = 81take the square root of both sides.x = ±9since the length cannot be a negative number, x = 9 m By Pythagoras, if the sum of squares of two sides is equal to the square of the largest side then the triangle is said to be right-angled


Here given sides of the triangle are 7, 24 and 25


Largest side = 25


252 = 625  Sum of squares of remaining two sides 72 + 242 = 49 + 576 = 625


∴ 72 + 242 = 252 Therefore triangle having sides 7,24 and 25 is a right-angled triangle let abc be the trianglewhere ab=(a-1)2 cmbc=2√a cmca=(a+1) cmab2=(a-1)2=a2+1-2abc2=(2√a)2=4a2ca2=(a+1)2=a2+1+2ahence ab2+bc2=ac2so Δabc is a right angles triangle at b let a, b, c be the sides of any triangle abc. then by applying the sine rule, we get⇒c = k sin csimilarly, b = k sin band a = k sin ahere we will consider rhs, so we getrhs = (b2 – c2) sin asubstituting corresponding values in the above equation, we get⇒ = [( k sin b)2 - ( k sin c)2] sin a⇒ = k2(sin2 b – sin2 c )sin a……….(ii)but,substituting the above values in equation (ii), we get⇒ = k2(sin(b + c) sin(b - c)) sin abut a + b + c = π ⇒ b + c = π –a, so the above equation becomes,⇒ = k2(sin(π –a) sin(b - c))sin abut sin (π - θ) = sin θ⇒ = k2(sin(a) sin(b - c))sin arearranging the above equation we get⇒ = (k sin(a))( sin(b - c))(k sin a)from sine rule, a = k sin a, so the above equation becomes,⇒ = a2 sin(b - c) = rhshence proved 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 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 a. dimensions of h = [l]dimensions of s = [mt-2]dimensions of ρ = [ml-3]dimensions of r = [l]dimensions of g = [lt-2]checking the r.h.sdimensionally= = [m0l1t0]a. is dimensionally correctb. dimensions of v = [lt-1]dimensions of p = [ml-1t-2]dimensions of ρ = [ml-3]checking the r.h.sdimensionally= b. is dimensionally correctc. dimensions of v = [l3]dimensions of p = [ml-1t-2]dimensions of r4 = [l4]dimensions of t = [t]dimensions of l = [l]dimensions of η = [ml-1t-1]checking the r.h.sdimensionally= =[l3]c. is dimensionally correctd. dimensions of v = [t-1]dimensions of m = [m]dimensions of g = [lt-2]dimensions of l = [l]dimensions of i = [ml2]checking the r.h.sdimensionally= d. is dimensionally correct consider the parallelogram abcd with diagonals ac and bd as shown and they intersect at right angles at oa parallelogram is a square if its adjacent angles are 90° and adjacent sides are equalso we have to prove that adjacent sides of given parallelogram are equal and adjacent angles are 90° to prove given parallelogram is a squareconsider δbad and δcdaac = bd … givenab = dc … opposite sides of a parallelogramad is the common sidetherefore, δbad ≅ δcda … sss test for congruency⇒ ∠bad = ∠cda …corresponding angles of congruent triangles …(i)as it is given that abcd is parallelogram⇒ ∠bad + ∠cda = 180° … sum of adjacent angles of a parallelogram is 180°using equation (i)⇒ ∠cda + ∠cda = 180°⇒ 2 × ∠cda = 180°⇒ ∠cda = 90°⇒ ∠bad = 90°thus, adjacent angles are 90° … (iii)consider δaob and δaod∠aob = ∠aod … both 90° because given that diagonals intersect at right anglesod = ob … diagonals of a parallelogram bisect each otherao is the common sidetherefore, δaob ≅ δaod … sas test for congruency⇒ ab = ad … corresponding sides of congruent trianglesthus, adjacent sides are equal … (iv)from statements (iii) and (iv) we can conclude that parallelogram abcd is a squaretherefore, if in a parallelogram, the diagonals are equal in lengths and intersect at right angles, the parallelogram will be a square. 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. 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! ;)