Quantitative Aptitude Ques 626
Question: 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
Options:
A) 10 cm
B) 12 cm
C) 6 cm
D) 8 cm
Show Answer
Answer:
Correct Answer: A
Solution:
- From figure, CE = ED = 8 cm, EB = 4 cm, OB =? Join OD. Then, OD = OB = Radius = x In $ \Delta OED, $ $ OD^{2}=OE^{2}+DE^{2} $ $ ={{(OB-EB)}^{2}}+DE^{2} $ $ x^{2}={{(x-EB)}^{2}}+DE^{2} $
$ \Rightarrow $ $ x^{2}={{(x-4)}^{2}}+8^{2} $
$ \Rightarrow $ $ x^{2}=x^{2}+16-8x+64 $
$ \Rightarrow $ $ 8x=80 $
$ \Rightarrow $ $ x=10 $
BETA
