Question
In the given figure, a circle with centre $O$ is given in which a diameter $AB$ bisects the chord $CD$ at a point E such that $CE = ED = 8cm$ and$ EB = 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 $EB = 4\ cm$ Join $OC$.
Let $OC = OB = rcm$ [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 $EB = 4\ cm$ Join $OC$.
Let $OC = OB = rcm$ [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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