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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