In $ \triangle \mathrm{ABC}, \angle \mathrm{B}=90^{\circ} $ and $ \mathrm{BM} $ is an altitude. If $ \mathrm{BM}=10 $ and $ \mathrm{CM}=5 $, find the perimeter of $ \triangle \mathrm{ABC} $.

Given:

In \( \triangle \mathrm{ABC}, \angle \mathrm{B}=90^{\circ} \) and \( \mathrm{BM} \) is an altitude.

\( \mathrm{BM}=10 \) and \( \mathrm{CM}=5 \)

To do:

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

Solution:

Let $AM=x$

In $\triangle BMC$,

By Pythagoras theorem,

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

$=10^2+5^2$

$=100+25$

$BC=\sqrt{125}$

Similarly,

In $\triangle ABM$,

By Pythagoras theorem,

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

$=x^2+10^2

$AB=\sqrt{x^2+100}$ 

In $\triangle ABC$,

By Pythagoras theorem,

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

$(x+5)^2=(\sqrt{x^2+100})^2+(\sqrt{125})^2$

$x^2+25+10x=x^2+100+125$

$10x=225-25$

$x=\frac{200}{10}$

$x=20$

$\Rightarrow AB=\sqrt{20^2+100}$

$=\sqrt{400+100}$

$=\sqrt{500}$

$=10\sqrt5$

$\Rightarrow BC=\sqrt{125}=5\sqrt5$

$\Rightarrow AC=20+5=25$

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

$=20+10\sqrt5+25$

$=45+10\sqrt{5}$ 

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Updated on: 10-Oct-2022

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