Question

The figure given below shows a circle with centre O in which diameter AB bisects the chord CD at point E. If CD =16 cm and EB = 4 cm, then find the radius of the circle.

Answer 1

Let OE = x cm.
Given, EB = 4 cm.
Then, the radius of the circle will be OB = OE + EB = (x+4) cm
$\Rightarrow OD=(x+4)cm$
Also, it is given that diameter AB bisects the chord CD at point E. Hence, $CE=ED=\frac{16}{2}cm=8cm$.
We know that if a line drawn from the centre of a circle bisects the chord, then the line is perpendicular to that chord.
i.e., $OE\perp ED$
Then, by using Pythagoras' theorem in $\Delta OED$, we get
$(4+x{)}^2=x^2+8^2$
$\Rightarrow 16+x^2+8x=x^2+64$
$\Rightarrow x=6$
$\therefore \text{Radius of circle}=x+4=6+4=10cm$