Question

In the given figure, $O$ is the centre of a circle of radius $r$ cm, $OP$ and $OQ$ are perpendiculars to $PQ=1cm$, If $AB\parallel CD,AB=6cm$ and $CD=8cm$, then determine r

Answer 1

In $\Delta AOP$, by Pythagoras theorem

$r^2=3^2+(1+x{)}^2$....(1)

In $\Delta COQ,r^2=4^2+x^2$...(2)

$e{q}^n\left(1\right)-e{q}^n\left(2\right)O=9-16+(1+x{)}^2-x^2$

$\Rightarrow 2x=6\to x=3$

By, eqn (1) $r^2=9+16\Rightarrow r^2=25\Rightarrow r=5cm$