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.

Answer 1

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.