Question

Calculate the area of the quadrilateral PQRS shown in the adjoining figure, it being given that PR = 8 cm, RQ = 17 cm, $\angle RPQ={90}^{\circ }$, RS $=$ 6 cm and $\angle PRS={90}^{\circ }$.

Answer 1

In right-angled triangle $PQR,$

$Q{R}^2=P{Q}^2+P{R}^2$ [by Pythagoras theorem]

$\Rightarrow P{R}^2=Q{R}^2-P{Q}^2$

$\Rightarrow PR=\sqrt{{17}^2-8^2}$

$=15$

Now, $ar\left(\Delta PQR\right)=\frac{1}{2}\times PQ\times QR$

$=\frac{1}{2}\times 8\times 15=60c{m}^2$

And $ar\left(\Delta PRS\right)=\frac{1}{2}\times PR\times SR$

$=\frac{1}{2}\times 6\times 15=45c{m}^2$

Hence, $ar\left(PQRS\right)=ar\left(PQR\right)+ar\left(PRS\right)$

$=60+45$

$=105c{m}^2$