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 $=AB+BC+AC$

$=15+20+25$

$=60$

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Updated on: 10-Oct-2022

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