Question #013ce

1 Answer
Jan 27, 2018

As proved below

Explanation:

Assumed (a2) / (b2)a2b2 as a^2 / b^2a2b2

Given : Sides are in the ratio a / bab

Let a1, a2, a3 be the sides of triangle a & b1, b2, b3 be the sides of triangle b

Since the two triangles are similar, corresponding angles of the two triangles are equal.

Area of a triangle A_T = (1/2) a * h = (1/2) a b sin CAT=(12)ah=(12)absinC

Since /_C is equal in both the triangles,

A_a / A_b =( cancel(1/2)a1* a2 cancelsin (A3)) / (cancel(1/2) b1* b2cancelsin(B3)) = (a1 * a2 ) / (b1 * b2), as sin (A3) = sin (B3)

A_a / A_b = ((a1)/(b1)) * ((a2) / (b2))

We know (a1) / (b1) = (a2) / (b2) = a/b

Hence A_a / A_b = (a/b) * (a/b) = a^2 / b^2