# 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})^2x^2+25+10x=x^2+100+12510x=225-25x=\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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