Two isosceles triangles have equal vertical angles and their areas are in the ratio 36 : 25. Find the ratio of their corresponding heights.
Solution
Given: AB = AC, PQ = PQ and ∠A = ∠P
And, AD and PS are altitudes
And, `("Area"(triangleABC))/("Area"(trianglePQR))=36/25` .........(i)
To find `"AD"/"PS"`
Proof: Since, AB = AC and PQ = PR
Then, `"AB"/"AC"=1` and `"PQ"/"PR"=1`
`therefore"AB"/"AC"="PQ"/"PR"`
`rArr"AB"/"PQ"="AC"/"PR"` ........(ii)
In ΔABC and ΔPQR
∠A = ∠P [Given]
`"AB"/"PQ"="AC"/"PR"` [From (2)]
Then, ΔABC ~ ΔPQR [By SAS similarity]
`therefore("Area"(triangleABC))/("Area"(trianglePQR))="AB"^2/"PQ"^2` .....(iii) [By area of similar triangle theorem]
Compare equation (i) and (iii)
`"AB"^2/"PQ"^2=36/25`
`"AB"/"PQ"=6/5` ..........(iv)
In ΔABD and ΔPQS
∠B = ∠Q [ΔABC ~ ΔPQR]
∠ADB = ∠PSQ [Each 90°]
Then, ΔABD ~ ΔPQS [By AA similarity]
`therefore"AB"/"PQ"="AD"/"PS"`
`rArr6/5="AD"/"PS"` [From (iv)]
