In $ \triangle \mathrm{ABC}, \angle \mathrm{B}=90^{\circ} $ and $ \mathrm{BM} $ is an altitude. If $ \mathrm{BM}=\sqrt{30} $ and $ \mathrm{CM}=3 $, find $ \mathrm{AC} $.

Given:

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

\( \mathrm{BM}=\sqrt{30} \) and \( \mathrm{CM}=3 \)

To do:

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

Solution:

We know that,

In a right-angled triangle, the altitude on the hypotenuse is equal to the geometric mean of line segments formed by altitude on the hypotenuse.

Therefore,

$BM^2=AM \times CM$

Let $AM=x$

This implies,

$(\sqrt{30})^2=x \times 3$

$30=3x$

$x=\frac{30}{3}$

$x=10$

Therefore,

$AC=AM+CM$

$=10+3$

$=13$

Hence, \( \mathrm{AC}=13 \).

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Updated on: 10-Oct-2022

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