The figure given below shows a circle with centre O in which diameter AB bisects the chord CD at point E. If CD =16 cm and EB = 4 cm, then find the radius of the circle.
Let OE=x cm.
Given, EB=4 cm.
Then, the radius of the circle will be OB=OE+EB=(x+4) cm
⟹OD=(x+4) cm
Also, it is given that diameter AB bisects the chord CD at point E. Hence, CE=ED=162=8 cm.
We know that if a line drawn from the centre of a circle bisects the chord, then the line is perpendicular to that chord.
i.e., OE⊥ED
Then, by using Pythagoras' theorem in ΔOED, we get
(4+x)2=x2+82
⟹16+x2+8x=x2+64
⟹x=6
∴Radius of circle=x+4=6+4=10 cm