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 = 6 cm

And, PS = 9 cm

By area of similar triangle theorem

`("Area"(triangleABC))/("Area"(trianglePQR))="AB"^2/"PQ"^2`              ........(i)

In ΔABD and ΔPQS

∠B = ∠Q [ΔABC ~ ΔPQR]

∠ADB = ∠PSQ [Each 90°]

Then, ΔABD ~ ΔPQS [By AA similarity]

`therefore"AB"/"PQ"="AD"/"PS"`               [Corresponding parts of similar Δ are proportional]

`rArr"AB"/"PQ"=6/9`

`rArr"AB"/"PQ"=2/3`              ......(ii)

Compare equations (i) and (ii)

`("Area"(triangleABC))/("Area"(trianglePQR))=(2/3)^2=4/9`