In $ \triangle \mathrm{ABC}, \angle \mathrm{B}=90^{\circ} $ and $ \mathrm{BE} $ is a median. If $ A B=15 $ and $ B E=8.5 $, find $ B C $.

Given:

In \( \triangle \mathrm{ABC}, \angle \mathrm{B}=90^{\circ} \) and \( \mathrm{BE} \) is a median.

\( A B=15 \) and \( B E=8.5 \),

To do:

We have to find \( B C \).

Solution:

We know that,

In a right-angled triangle, the median bisecting the hypotenuse is half of the hypotenuse.

Therefore,

$BE=\frac{AC}{2}$

$8.5=\frac{AC}{2}$

$AC=8.5\times2$

$AC=17$

In right-angled triangle $\mathrm{ABC}$, by Pythagoras theorem,

$AC^2=AB^2+BC^2$

$17^2=15^2+BC^2$

$BC^2=289-225$

$BC=\sqrt{64}$

$BC=8$

Hence, $BC=8$.

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Updated on: 10-Oct-2022

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