In the given figure, the diameter CD of a circle with centre O is perpendicular to the chord AB.

If AB = 8 cm and CM = 2 cm, find the radius of the circle.

Given: Diameter CD of the circle is perpendicular to the chord AB. AB = 8 cm and CM = 2 cm. O is the center of the circle. r is the radius of the circle.

To prove: Radius of the circle (r).

Construction: Join OA.

Proof: Since CD is perpendicular to AB, it bisects the chord.

Thus AM = MB = $\dfrac{AB}{2} = \dfrac{8}{2}$ = 4 cm.

OM = r - CM = r - 2 cm

In Δ OAM, ∠M = 90°

Using Pythagoras theorem,

∴ OA² = OM² + AM²

⇒ r² = (r - 2)² + 4²

⇒ r² = r² + 4 - 4r + 16

⇒ 0 = 16 + 4 - 4r

⇒ 4r = 20

⇒ r = $\dfrac{20}{4}$

⇒ r = 5 cm

Hence, the radius of the circle = 5 cm.