question_answer

In the given figure, O is the centre of a circle and diameter AB bisects the chord CD at a point E such that CE = ED = 8 cm and EB = 4 cm. The radius of the circle is

A) 10 cm  

B) 12 cm

C) 6 cm                            

D) 8 cm

Correct Answer: A

Solution :

From figure, CE = ED = 8 cm, EB = 4 cm, OB =? Join OD.

Then, OD = OB = Radius = x

In \[\Delta OED,\]\[O{{D}^{2}}=O{{E}^{2}}+D{{E}^{2}}\]

\[={{(OB\,\,-EB)}^{2}}+D{{E}^{2}}\]

\[{{x}^{2}}={{(x-EB)}^{2}}+D{{E}^{2}}\]

\[\Rightarrow \]   \[{{x}^{2}}={{(x-4)}^{2}}+{{8}^{2}}\]

\[\Rightarrow \]   \[{{x}^{2}}={{x}^{2}}+16-8x+64\]

\[\Rightarrow \]   \[8x=80\]\[\Rightarrow \]\[x=10\]