Question
The figure given below shows a circle with center $O$ in which diameter $AB$ bisects the chord $CD$ at point $E$. If $CE = ED = 8 \ cm$ and $EB = 4 \ cm,$
find the radius of the circle.
find the radius of the circle.
✓
Answer
Let the radius of the circle be $r \ cm.$
$\therefore OE = OB - EB = r - 4$
Join $OC.$
In right $\triangle OEC,$
$OC^2 = OE^2 + CE^2$
$\Rightarrow r^2 = ( r - 4 )^2 + (8)^2$
$\Rightarrow r^2 = r^2 - 8r + 16 + 64$
$\Rightarrow 8r = 80$
$\therefore r = 10 \ cm$
Hence, radius of the circle is $10 \ cm.$
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