Question

The areas of two similar triangles are 25 cm² and 36 cm² respectively. If the altitude of the first triangle is 2.4 cm, find the corresponding altitude of the other.

Answer 1

We have,

ΔABC ~ ΔPQR

Area (ΔABC) = 25 cm²

Area (ΔPQR) = 36 cm²

AD = 2.4 cm

And AD and PS are the altitudes

To find: PS

Proof: Since, ΔABC ~ ΔPQR

Then, by area of similar triangle theorem

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

`rArr25/36="AB"^2/"PQ"^2`

`rArr5/6="AB"/"PQ"`                 .........(i)

In ΔABD and ΔPQS

∠B = ∠Q [Δ ABC ~ ΔPQR]

∠ADB ~ ∠PSQ [Each 90°]

Then, ΔABD ~ ΔPQS [By AA similarity]

`therefore"AB"/"PS"="AD"/"PS"`           ......(ii)   [Corresponding parts of similar Δ are proportional]

Compare (i) and (ii)

`"AD"/"PS"=5/6`

`rArr2.4/"PS"=5/6`

`"PS"=(2.4xx6)/5=2.88` cm