In $ \triangle \mathrm{XYZ}, \angle \mathrm{Y}=90^{\circ} $ and $ \mathrm{YM} $ is an altitude. If $ \mathrm{XM}=\sqrt{12} $ and $ \mathrm{ZM}=\sqrt{3} $, find $ \mathrm{YM} $.

Given:

In \( \triangle \mathrm{XYZ}, \angle \mathrm{Y}=90^{\circ} \) and \( \mathrm{YM} \) is an altitude.

\( \mathrm{XM}=\sqrt{12} \) and \( \mathrm{ZM}=\sqrt{3} \)

To do:

We have to find \( \mathrm{YM} \).

Solution:

In $\triangle XYZ$,

By Pythagoras theorem,

$XZ^2 =XY^2+YZ^2$

$(\sqrt{12}+\sqrt{3})^2=XY^2+YZ^2$

$(\sqrt{12})^2+(\sqrt{3})^2+2\times\sqrt{12}\times\sqrt{3}=XY^2+YZ^2$

$12+3+2\times\sqrt{36}=XY^2+YZ^2$

$15+2\times6=XY^2+YZ^2$

$15+12=XY^2+YZ^2$

$27=XY^2+YZ^2$.........(i)

Similarly,

In $\triangle XYM$,

By Pythagoras theorem,

$XY^2 =XM^2+YM^2$

$XY^2=(\sqrt{12})^2+YM^2$

$XY^2=12+YM^2$......(ii) 

In $\triangle YMZ$,

By Pythagoras theorem,

$YZ^2 =MZ^2+YM^2$

$YZ^2=(\sqrt{3})^2+YM^2$

$YZ^2=3+YM^2$......(iii)

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

$27=12+YM^2+3+YM^2$

$27-15=YM^2+YM^2$

$12=2YM^2$

$YM^2=6$

$YM=\sqrt{6}$

Hence, $YM=\sqrt{6}$

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Updated on: 10-Oct-2022

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