Question

In figure $\angle QPR={90}^{\circ },segPM\perp segQR$ and $PM=10,QM=8,$ find area of $\Delta$ $PQR$.

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, $8^2+2\left({10}^2\right)+M{R}^2=8^2+16MR+M{R}^2$

or, $16MR=200$

or, $MR=12.5$.

Now $QR=8+12.5=20.5$.

$\therefore \Delta$ $PQR=$$\frac{1}{2}\times 20.5\times 10=102.5$ sq.unit