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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 \( \frac{4}{3} \).
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