In $ \triangle \mathrm{ABC}, \angle \mathrm{B}=90^{\circ} $ and $ \mathrm{BM} $ is an altitude. If $ \mathrm{AB}=2 \sqrt{10} $ and $ \mathrm{AM}=5 $, find $ \mathrm{CM} $.

Given:

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

\( \mathrm{AB}=2 \sqrt{10} \) and \( \mathrm{AM}=5 \)

To do:

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

Solution:

In right angled triangle $ABM$, by Pythagoras theorem,

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

$(2\sqrt{10})^2=5^2+BM^2$

$4\times10=25+BM^2$

$BM^2=40-25$

$BM=\sqrt{15}$

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.

$BM^2=AM \times CM$

Let $CM=x$

This implies,

$(\sqrt{15})^2=5 \times x$

$15=5x$

$x=\frac{15}{5}$

$x=3$

Hence, \( \mathrm{CM}=3 \).