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

Given,

AB = AC = $12\sqrt{5}$ cm

In an isosceles triangle, the perpendicular from a vertex between equal sides bisects the opposite side.

Thus,

L is the mid-point of BC.

∴ BL = $\dfrac{BC}{2} = \dfrac{24}{2}$ = 12 cm.

AL is perpendicular bisector of BC and perpendicular from center bisects the chord.

Thus, centre of the circle O lies on AL. Let radius of circle OB be r.

In right-angled triangle ALB,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ AB² = AL² + BL²

⇒ ($12\sqrt{5}$)² = AL² + 12²

⇒ 720 = AL² + 144

⇒ AL² = 720 - 144

⇒ AL² = 576

⇒ AL = $\sqrt{576}$ = 24 cm

OL = AL - AO = 24 - r

In right-angled triangle OBL,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OB² = OL² + BL²

⇒ r² = (24 - r)² + 12²

⇒ r² = r² - 48r + 576 + 144

⇒ 48r = 720

⇒ r = $\dfrac{720}{48}$ = 15 cm

Hence, radius of the circle = 15 cm.