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**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,

OC^{2} = OE^{2} + CE^{2}

⇒ r^{2} = ( r - 4 )^{2} + (8)^{2 }

⇒ r^{2} = r^{2} - 8r + 16 + 64

⇒ 8r = 80

∴ r = 10 cm

Hence, radius of the circle is 10 cm.

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