Question

$PQRS$ is a cyclic quadrilateral. $PR$ is the diameter of the circle. If $PQ=7cm,QR=6cm$ and $RS=cm$, then $PS=$:

• A. $7cm$
• B. $8cm$
• C. $9cm$
• D. $10cm$

Answer 1

The correct option is C $9cm$

As given in the figure, $\square PQRS$ is cyclic quadrilateral and PR is diagonal of quadrilateral which is diameter of circle.

Thus, by property of cyclic quadrilateral, circle will pass through all the vertices of quadrilateral as shown in figure.

Now, PR is diameter of circle. Thus, $\angle PQR$ is angle in semi-circle.

$\therefore \angle PQR={90}^0$ (Property of angle in semi-circle)

It is given that $PQ=7$ cm and $QR=6$ cm

In $△PQR$, by Pythagoras theorem,

${PR}^2={PQ}^2+{QR}^2$

$\therefore {PR}^2={\left(7\right)}^2+{\left(6\right)}^2$

$\therefore {PR}^2=85$

$\therefore PR=\sqrt{85}$ cm

Now, $\angle PSR$ is also angle in semi-circle

$\therefore \angle PSR={90}^0$

Also it is given that $RS=2$ cm

In $△PSR$, by Pythagoras theorem,

${PR}^2={PS}^2+{SR}^2$

$\therefore 85={PS}^2+{\left(2\right)}^2$

$\therefore 85={PS}^2+4$

$\therefore {PS}^2=85-4=81$

Taking square roots on both sides, we get,

$PS=\sqrt{81}$

$\therefore PS=9$ cm

Thus, answer is option (C)