Question
In the given figure, a circle with centre $O$ is given in which a diameter $A B$ bisects the chord $C D$ at a point $E$ such that $C E=E D=8 cm$ and $E B=4 cm$. Find the radius of the circle.
✓
Answer
$AB$ is the diameter of the circle with centre $O$, which bisects the chord $CD$ at point $E$.
Given: $CE = ED = 8\ cm$ and $EB = 4cm$ Join $OC.$
Let $OC = OB =$ $r\ cm$ [Radii of a circle]
Then $OE$ $= (r - 4)\ cm$ Now, in right angled $\triangle\text{OEC},$
we have: $OC^2 = OE^2 + EC^2$[Pythagoras theorem]
$\Rightarrow r^2 = (r - 4)^2 + 8^2$
$\Rightarrow r^2 = r^2 - 8r + 16 + 64$
$\Rightarrow r^2 = r^2 + 8r = 80$
$\Rightarrow 8r = 80$
$\Rightarrow\ \text{r}=\Big(\frac{80}{8}\Big)\text{cm}=10\text{cm}$
$\Rightarrow r = 10\ cm$
Hence, the required radius of the circle is $10\ cm$.
Given: $CE = ED = 8\ cm$ and $EB = 4cm$ Join $OC.$
Let $OC = OB =$ $r\ cm$ [Radii of a circle]
Then $OE$ $= (r - 4)\ cm$ Now, in right angled $\triangle\text{OEC},$
we have: $OC^2 = OE^2 + EC^2$[Pythagoras theorem]
$\Rightarrow r^2 = (r - 4)^2 + 8^2$
$\Rightarrow r^2 = r^2 - 8r + 16 + 64$
$\Rightarrow r^2 = r^2 + 8r = 80$
$\Rightarrow 8r = 80$
$\Rightarrow\ \text{r}=\Big(\frac{80}{8}\Big)\text{cm}=10\text{cm}$
$\Rightarrow r = 10\ cm$
Hence, the required radius of the circle is $10\ cm$.
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