Question

The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively. Find the ratio of their areas.

Answer 1

We have, 

ΔABC ~ ΔPQR 

AD = 6cm 

PS = 9cm 

By area of similar triangle theorem 

Area of ΔABC/Area of ΔPQR = AB² /PQ²…………(i) 

In ΔABD and ΔPQS

∠B = ∠Q (ΔABC ~ ΔPQS)

∠ADB = ∠PSQ (Each 90°)

Then, ΔABD ~ ΔPQS (By AA Similarity) 

So, AB/PQ = AD/PS (Corresponding parts of similar Δ are proportional) 

Or, AB/PQ = 6/9 

Or, AB/PQ = 2/3 ……………….(ii) 

Compare equation (i) and (ii) 

Area of ΔABC/Area of ΔPQR = (2/3)² = 4/9