Question

In figure $\angle QPR={90}^{\circ },segPM\perp segQR$ and $Q-M-R,PM=10,QM=8,$ find $QR$.

Answer 1

We have from the figure $\Delta$ PMQ is right-angled triangle.

Then $P{Q}^2={10}^2+8^2$......(1)

Also from right-angled triangle $\Delta$ PMR we have

$P{R}^2=M{R}^2+{10}^2$.......(2).

Now from right-angled triangle $\Delta$ PQR we get,

$P{R}^2+P{Q}^2=(8+MR{)}^2$

or, $8^2+{2.10}^2+M{R}^2=8^2+16MR+M{R}^2$

or, $16MR=200$

or, $MR=12.5$.

$QR=12.5+8=20.5$