The figure shows a circle with centre O, in which the diameter AB bisects the chord CD at point E. If CE = ED = 8 cm and EB = 4 cm. Find the radius of the circle.
[3 Marks]
Join OD. Let radius of the circle be 'r' cm.
OB = OD = r cm
[1 Mark]
EB = 4 cm (given)
∴ OE = (r - 4) cm
In △OED, ∠OED=90o
Using Pythagoras' theorem in △OED, we get
OD2 = OE2+ED2
[1 Mark]
⇒ r2=(r−4)2+82
⇒ r2=(r2−8r+16)+64
⇒ r=10 cm
[1 Mark]