The figure given below, shows a circle with centre O in which diameter AB bisects the chord CD at point E. If CE = ED = 8 cm and EB = 4cm, find the radius of the circle.
Let the radius of the circle be r cm.
∴ OE = OB- EB = r-4
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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