Question

The figure, given below, shows the circle with centre O in which diameter AB bisects the chord CD at point E. If $CE=ED=8$ and $EB=4$, find the radius of the circle.

Answer 1

Let the radius of the circle be r cm

$OE=OB-EB=r-4$

Let us join OC

In $\Delta OEC$

$(OC{)}^2=(OE{)}^2+(CE{)}^2$

$\Rightarrow r^2=(r-4{)}^2+(8{)}^2$

$\Rightarrow r^2=r^2-8r+16+69$

$\Rightarrow 8r=80$

$\Rightarrow r=10cm$