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

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by priyanka26

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by priyanka26

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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

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