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