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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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