ABC is an isosceles triangle inscribed in a circle. If AB = AC = $12\sqrt{5}$ cm and BC = 24 cm, find the radius of the circle.

From figure,

OA = radius = r cm.

BD = DC = $\dfrac{24}{2}$ = 12 cm (As perpendicular to a chord from the center of the circle bisects it)

In right angle triangle ABD,

$\Rightarrow AB^2 = BD^2 + AD^2 \\[1em] \Rightarrow (12\sqrt{5})^2 = 12^2 + AD^2 \\[1em] \Rightarrow 720 = 144 + AD^2 \\[1em] \Rightarrow AD^2 = 720 - 144 \\[1em] \Rightarrow AD^2 = 576 \\[1em] \Rightarrow AD = \sqrt{576} = 24 \text{ cm}.$

OD = AD - OA = (24 - r) cm.

In right angle triangle OBD,

⇒ OB = radius = r cm.

⇒ OB² = OD² + BD²

⇒ r² = (24 - r)² + 12²

⇒ r² = 576 + r² - 48r + 144

⇒ r² - r² + 48r = 720

⇒ 48r = 720

⇒ r = $\dfrac{720}{48}$ = 15 cm.

Hence, radius = 15 cm.