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 = 4cm. Find the radius of the circle.
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Answer
AB is the diameter of the circle with centre O, which bisects the chord CD at point E.
Given: CE = ED = 8cm and EB = 4cm
Join OC.
Let OC = OB = rcm [Radii of a circle]
Then OE = (r - 4)cm
Now, in right angled $\triangle\text{OEC},$ we have:
OC2 = OE2 + EC2 [Pythagoras theorem]
⇒ r2 = (r - 4)2 + 82
⇒ r2 = r2 - 8r + 16 + 64
⇒ r2 = r2 + 8r = 80
⇒ 8r = 80
$\Rightarrow\ \text{r}=\Big(\frac{80}{8}\Big)\text{cm}=10\text{cm}$
⇒ r = 10cm
Hence, the required radius of the circle is 10cm.
Given: CE = ED = 8cm and EB = 4cm
Join OC.
Let OC = OB = rcm [Radii of a circle]
Then OE = (r - 4)cm
Now, in right angled $\triangle\text{OEC},$ we have:
OC2 = OE2 + EC2 [Pythagoras theorem]
⇒ r2 = (r - 4)2 + 82
⇒ r2 = r2 - 8r + 16 + 64
⇒ r2 = r2 + 8r = 80
⇒ 8r = 80
$\Rightarrow\ \text{r}=\Big(\frac{80}{8}\Big)\text{cm}=10\text{cm}$
⇒ r = 10cm
Hence, the required radius of the circle is 10cm.
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