In the figure, line AB is perpendicular to lines PQ and RS. If AB = 7 cm, RS = 6 cm, and PQ = 8 cm, then find the radius of the circle.

Given: AB = 7 cm

RS = 6 cm

PQ = 8 cm

Let r be the radius of the circle.

Since perpendicular from the centre of the circle to a chord bisects the chord.

$\therefore AR=\frac{RS}{2}=3 cm \mathrm{and} BP=\frac{PQ}{2}=4 cm$

In $∆BOP,\angle OBP={90}^0$

$\therefore O{P}^2=O{B}^2+B{P}^2$       [By Pythagoras theorem]

$\Rightarrow r^2=x^2+{\left(4\right)}^2$

$\Rightarrow r^2=x^2+16$   …..(i)

Similarly, in $∆AOR,\angle OAR={90}^0$

$\therefore O{R}^2=O{A}^2+A{R}^2$      [By Pythagoras theorem]

$\Rightarrow r^2={(7-x)}^2+3^2$

$\Rightarrow r^2={(7-x)}^2+9$     ….(ii)

From equations (i) and (ii), we get,

$\Rightarrow x^2+16={(7-x)}^2+9$

$\Rightarrow x^2+16=7^2-2\times 7\times x+x^2+9$

$\Rightarrow 14x=49-16+9$

$\Rightarrow x=3$

When substituting the value of $x$ in equation (i), we get,

$\Rightarrow r^2=9+16=25$

 $\Rightarrow r=5 cm$

Hence, the radius of the circle is $5 cm$.