Prove that areas of similar triangles have the same ratio as the squares of the radii of their circum circle

1
by priyanka26

2015-04-26T22:19:15+05:30

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
R1 = radius of circum circle. of ABC with sides AB = c, BC = a and CA = b.
s1 = semiperimeter = (a+b+c)/2

Δ1 = area of the triangle ABC = √[ s1 (s1-a) (s1-b) (s1-c) ]
R1 = a b c / [4 Δ1]
Δ1 = a b c / [4 R1]

Let ABC and DEF be similar triangles.  Then their sides are proportional and their interior angles are same.
DE / AB = EF / BC = FD / CA = k
Without the loss of generality, let DE=ka, EF=kb, FD=kc.

s2 = (ka+kb+kc)/2 = k(a+b+c)/2 = k s1
Δ2 = area of triangle DEF = √[ s2 (s2 - ka)(s2 - kb) (s2 - kc) ]
= √ [ k s1 (ks1 - ka) (k s1 - k b) (ks1 - kc) ]
= √ [ k⁴ s1(s1-a) (s1 - b) (s1 - c)
= k² Δ1

R2 = radius of circum circle of triangle DEF = (ka) (kb) (kc) / [4 Δ2]
= k³ abc / [ 4 k² Δ1 ]
= k  R1

(R2 / R1)² = k² = Δ2 /Δ1