In $ \triangle \mathrm{PQR}, \angle Q=90^{\circ} $ and $ \mathrm{QM} $ is a median. If $ \mathrm{PQ}=20 $ and $ Q R=21 $, find $ Q M $.

Given:

In \( \triangle \mathrm{PQR}, \angle Q=90^{\circ} \) and \( \mathrm{QM} \) is a median. If \( \mathrm{PQ}=20 \) and \( Q R=21 \).

To do:

We have to find \( Q M \).

Solution:

In $\triangle PQR$,

By Pythagoras theorem,

$PR^2 =PQ^2+QR^2$

$=(20)^2+(21)^2$

$=400+441$

$=841$

$PR=\sqrt{841}=29$.......(i)

Similarly,

In $\triangle PQM$,

By Pythagoras theorem,

$PQ^2 =PM^2+QM^2$

$20^2=PM^2+QM^2$......(ii) 

In $\triangle QMR$,

By Pythagoras theorem,

$QR^2 =MR^2+QR^2$

$21^2=(29-PM)^2+QM^2$......(iii)

Subtracting (ii) from (iii)

$441-400=841+PM^2-58PM+QM^2-PM^2-QM^2$

$41=841-58PM$

$58PM=841-41$

$PM=\frac{800}{58}$

$PM=\frac{400}{29}$

Therefore,

$QM^2=400-(13.79)^2$

$=400-190.16$

$=209.84$

$QM=\sqrt{209.84}$

$QM=14.48$

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Updated on: 10-Oct-2022

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