An equilateral triangle is inscribed in a circle of radius 6 cm. Find its side.

Let ABC be an equilateral triangle inscribed in a circle of radius 6 cm. Let O be the centre of the circle.

Then, OA = OB = OC = 6 cm.

Let OD be perpendicular to O on side BC. Then, D is the mid-point of BC and OB and OC are bisectors of ∠B and ∠C, respectively.

∴ ∠OBD = 30°

In ∆OBD, right-angled at D, we have

∠OBD = 30° and OB = 6 cm.

cos ∠OBD = BD/OB

cos30°=BD/6

$BD=6\times \frac{\sqrt{3}}{2}=3\sqrt{3}$

now, BC = 2 BD = 2(3√3 )cm = 6 √3 cm.