Question

In the given figure, $PR$ is the diameter of the circle. $PQ=7$ cm, $QR=6$ cm and $RS=2$ cm. The perimeter of the cyclic quadrilateral $PQRS$ is

• A. $18$ cm
• B. $20\sqrt{2}$ cm
• C. $24$ cm
• D. $22\sqrt{3}$ cm

Answer 1

The correct option is C $24$ cm

In $△PQR$,
$\angle PQR={90}^{\circ }$ (Angle in a semi circle)
$P{R}^2=P{Q}^2+Q{R}^2$ (Pythagoras theorem)
$P{R}^2=7^2+6^2$
$PR=\sqrt{85}$
Now, In $△PRS$
$\angle PSR={90}^{\circ }$ (Angle is a semi circle)
$P{R}^2=P{S}^2+S{R}^2$
$85=P{S}^2+4$
$PS=\sqrt{81}=9$
Thus, perimeter of the quadrilateral = $PS+SR+QR+PQ$ = $9+2+6+7=24$ cm