Question

PQRS is a quadrilateral in which $PQ=5\sqrt{2}cm$, $QR=5\sqrt{2}cm$, $RS=8cm$, $SP=6cm$, $\angle PSR=\angle PQR={90}^{\circ }$. Then its area is

• A. $40.4c{m}^2$
• B. $46.4c{m}^2$
• C. $32.4c{m}^2$
• D. $49c{m}^2$

Answer 1

The correct option is D

$49c{m}^2$

Area of right angled triangle $\Delta PQR$
$=\frac{1}{2}\times Base\times Height=\frac{1}{2}\times 5\sqrt{2}\times 5\sqrt{2}=25c{m}^2$

Area of right angled triangle $\Delta PSR$
$=\frac{1}{2}\times 8\times 6=24c{m}^2$

Area of quadrilateral PQRS
= Area of $\Delta PQR+\Delta PSR=49c{m}^2$