In $ \triangle \mathrm{ABC}, \angle \mathrm{B}=90^{\circ} $ and $ \mathrm{BM} $ is an altitude. If $ A M=9 $ and $ C M=16 $, find the perimeter of $ \triangle \mathrm{ABC} $.

Given:

In \( \triangle \mathrm{ABC}, \angle \mathrm{B}=90^{\circ} \) and \( \mathrm{BM} \) is an altitude. If \( A M=9 \) and \( C M=16 \).

To do:

We have to find the perimeter of \( \triangle \mathrm{ABC} \).

Solution:

In $\triangle ABC$,

By Pythagoras theorem,

$AC^2 =AB^2+BC^2$

$(9+16)^2=AB^2+BC^2$

$(25)^2=AB^2+BC^2$

$625=AB^2+AC^2$.........(i)

Similarly,

In $\triangle ABM$,

By Pythagoras theorem,

$AB^2 =AM^2+BM^2$

$AB^2=9^2+BM^2$......(ii) 

In $\triangle BMC$,

By Pythagoras theorem,

$BC^2 =MC^2+BM^2$

$BC^2=(16)^2+BM^2$......(iii)

From (i), (ii) and (iii),

$625=81+BM^2+256+BM^2$

$625-337=2BM^2$

$288=2BM^2$

$BM^2=144$

$BM=\sqrt{144}=12$

Therefore,

$AB^2=81+144$

$=225$

$\Rightarrow AB=\sqrt{225}$

$=15$

$BC^2=256+144$

$=400$

$\Rightarrow BC=\sqrt{400}$

$=20$

Therefore, 

Perimeter of triangle ABC $=AB+BC+AC$

$=15+20+25$

$=60$