Quantitative Aptitude Ques 2443
Question: 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=4cm. $ The radius of the circle is
Options:
A) 10 cm
B) 12 cm
C) 6 cm
D) 8 cm
Show Answer
Answer:
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, $ OD^{2}=OE^{2}+ED^{2} $
$ \Rightarrow $ $ r^{2}={{(r-4)}^{2}}+8^{2} $
$ \Rightarrow $ 8r = 80
$ \therefore $ r = 10 cm
BETA
