Question

Question #013ce

Answer 1

As proved below

Explanation:

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

Given : Sides are in the ratio `a / b`

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 C`

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`