Question

The given figure shows a circle centred at O in which the diameter AB bisects the chord CD at point E. If CD = 16 𝑐𝑚 and EB = 4 𝑐𝑚, then find the radius of the circle.

Answer 1

Given: AB bisects the chord CD at point E, i.e., OE bisects the chord CD.
OE ⊥ CD (by theorem)
ED = CE = 16/2 cm = 8 cm
Let ‘$r$’ be the radius of the circle.
OD = OB = $r$ cm
OE = OB − EB = ($r$ − 4) cm
In △OED,
By Pythagoras theorem,
$O{D}^2=O{E}^2+E{D}^2$
$r^2=(r-4{)}^2+8^2$
∴ $r^2=r^2-8r+16+64$
∴ $8r=80$
∴ $r=10$ cm
Answer: radius = $10$ cm