In $ \triangle \mathrm{ABC}, \angle \mathrm{B}=90^{\circ} $ and $ \mathrm{BM} $ is a median. If $ \mathrm{AB}=15 $ and $ \mathrm{BC}=20 $, find $ \mathrm{BM} $.


Given:

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

\( \mathrm{AB}=15 \) and \( \mathrm{BC}=20 \)

To do:

We have to find \( B M \).

Solution:

In $\triangle ABC$,

By Pythagoras theorem,

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

$=(15)^2+(20)^2$

$=225+400$

$=625$

$AC=\sqrt{625}=25$.......(i)

Similarly,

In $\triangle ABM$,

By Pythagoras theorem,

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

$15^2=AM^2+BM^2$......(ii) 

In $\triangle BMC$,

By Pythagoras theorem,

$BC^2 =MC^2+BM^2$

$20^2=(25-AM)^2+BM^2$......(iii)

Subtracting (ii) from (iii)

$400-225=625+AM^2-50AM+BM^2-AM^2-BM^2$

$175=625-50AM$

$50AM=625-175$

$AM=\frac{450}{50}$

$AM=9$

Therefore,

$BM^2=225-(9)^2$

$=225-81$

$=144$

$BM=\sqrt{144}$

$BM=12$ 

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Updated on: 10-Oct-2022

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