Question

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

Answer 1

Let ΔABC and ΔPQR be similar triangles.

Let O₁ and R₁ be the circumcentre and the circumradius of ΔABC and O₂ and R₂ be the circumcentre and circumradius of ΔPQR.

We know that the angle subtended by a chord at the centre is twice the angle subtended by the same chord at any point on the remaining part of the circle.

∴ ∠ BO₁C = 2∠A and ∠QO₂R = 2∠P

Since ΔABC and ΔPQR are similar, ∠A = ∠ P.

Hence, ∠BO₁C = ∠QO₂R

In Δ BO₁C and Δ QO₂R, we have:

∴ Δ BO₁C ||| Δ QO₂R (By SAS postulate)

The ratio of the areas of similar triangles is equal to the ratio of the squares of their corresponding sides.