In the given figure, O is the centre of a circle and diameter AB bisects the chord CD at a point such that \[CE=ED=8\,cm\] and \[EB=4\text{ }cm.\]The radius of the circle is |
A) 10 cm
B) 12 cm
C) 6 cm
D) 8 cm
Correct Answer: A
Solution :
Let the radius of the circle be r cm. |
Then, OD = OB = r cm, \[OE=(r-4)\,cm\] |
and ED = 8 cm |
Now, \[O{{D}^{2}}=O{{E}^{2}}+E{{D}^{2}}\] |
\[\Rightarrow \] \[{{r}^{2}}={{(r-4)}^{2}}+{{8}^{2}}\] |
\[\Rightarrow \] 8r = 80 |
\[\therefore \] r = 10 cm |
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