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Question
The figure given below shows a circle with center O in which diameter AB bisects the chord CD at point E. If CE = ED = 8 cm and EB = 4 cm,
find the radius of the circle.
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Solution
Let the radius of the circle be r cm.
∴ OE = OB - EB = r - 4
Join OC.
In right ΔOEC,
OC2 = OE2 + CE2
⇒ r2 = ( r - 4 )2 + (8)2
⇒ r2 = r2 - 8r + 16 + 64
⇒ 8r = 80
∴ r = 10 cm
Hence, radius of the circle is 10 cm.
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