Question

CD is the diameter of the circle centered at O, which meets the chord AB at E. OE$\perp$AB and AB = 8 cm. If DE = 3cm , then the radius of the circle is

• A. $3\frac{1}{6}cm$
• B. $4\frac{1}{3}cm$
• C. $4\frac{1}{5}cm$
• D. $4\frac{1}{6}cm$

Answer 1

The correct option is D $4\frac{1}{6}cm$

Given, OE$\perp$AB and AB = 8 cm.
We know that the perpendicular from the centre of a circle to its chord bisects the chord.
Hence, AE = EB = 4 cm.

Let radius be $r$ cm $\Rightarrow OD=rcm$

$\therefore OE=(r-3)cm$

In$△OAE$, $\angle OEA={90}^o$.

Applying Pythagoras' theorem, we get

$O{A}^2=O{E}^2+A{E}^2$

$r^2=(r-3{)}^2+4^2$

$r^2=r^2+9-6r+16$

$\Rightarrow r=4\frac{1}{6}$ cm