SATHEE: Quantitative Aptitude Ques 626

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 $

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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:

Answer:

Solution:

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