#### Question

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

#### Solution

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`

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#### APPEARS IN

Solution for question: The Corresponding Altitudes of Two Similar Triangles Are 6 Cm and 9 Cm Respectively. Find the Ratio of Their Areas. concept: Areas of Similar Triangles. For the course CBSE