#### Question

The areas of two similar triangles are 100 cm^{2} and 49 cm^{2} respectively. If the altitude the bigger triangle is 5 cm, find the corresponding altitude of the other.

#### Solution

We have, ΔABC ~ ΔPQR

Area(ΔABC) = 100 cm^{2},

Area (ΔPQR) = 49 cm^{2}

AD = 5 cm

And AD and PS are the altitudes

By area of similar triangle theorem

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

`rArr100/49="AB"^2/"PQ"^2`

`rArr10/7="AB"/"PQ"` ............(i)

In ΔABD and ΔPQS

∠B = ∠Q [ΔABC ~ ΔPQR]

∠ADB = ∠PSQ [Each 90°]

Then, ΔABD ~ ΔPQS [By AA similarity]

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

Compare (i) and (ii)

`"AD"/"PS"=10/7`

`rArr5/"PS"=10/7`

`rArr"PS"=(5xx7)/10=3.5` cm