In $ \triangle \mathrm{ABC}, \angle \mathrm{B}=90^{\circ} $ and $ \mathrm{BM} $ is an altitude. If $ \mathrm{AM}=x+7, \mathrm{BM}=x+2 $ and $ \mathrm{CM}=x $, find
the value of $ x $.

Given:

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

\( \mathrm{AM}=x+7, \mathrm{BM}=x+2 \) and \( \mathrm{CM}=x \)

To do:

We have to find the value of \( x \).

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.

$BM^2=AM \times CM$

Let $CM=x$

This implies,

$(x+2)^2=(x+7) \times (x)$

$x^2+4+2\times x \times 2=x^2+7x$

$x^2+4+4x=x^2+7x$

$7x-4x=4$

$3x=4$

$x=\frac{4}{3}$

Hence, the value of $x$ is \( \frac{4}{3} \).