Question

The figure shows a circle with centre O, in which $OE\perp CD$. If CD = 8 cm and EB = 2 cm, then find the radius of the circle. [3 Marks]

Answer 1

Given, $OE\perp CD$ where $O$ is the centre of the circle.
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
So, $CE=ED=4$ cm
Join OD. Let radius of the circle be '$r$' cm.
OB = OD = $r$ cm [1 Mark]
EB = 2 cm (given)
$\therefore$ OE = ($r$ - 2) cm
In $△OED$, $\angle OED={90}^o$
Using Pythagoras' theorem in $△OED$, we get
$O{D}^2=O{E}^2+E{D}^2$ [1 Mark]
$\Rightarrow$ $r^2=(r-2{)}^2+4^2$
$\Rightarrow$ $r^2=(r^2-4r+4)+16$
$\Rightarrow$ $4r=20$
$\Rightarrow$ $r=5$ cm [1 Mark]