Question

Two isosceles triangles have equal vertical angles and their areas are in the ratio 36 : 25. Find the ratio of their corresponding heights.

Answer 1

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)]