An equilateral triangle of side 9 cm is inscribed in a circle. Find the radius of the circle.

Let ABC be equilateral triangle inscribed in circle.

AB = BC = AC

Draw AD ⊥ BC.

Since, in an equilateral triangle the perpendicular from a vertex bisects the opposite side.

Thus,

D is the mid-point of BC.

∴ BD = $\dfrac{BC}{2} = \dfrac{9}{2}$ = 4.5 cm.

Since, the chord BC is bisected at point D, and perpendicular from center bisects the chord.

Centre of the circle O lies on AD. Let radius of circle (OA) be r.

In right-angled triangle ADB,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ AB² = AD² + BD²

$\Rightarrow 9^2 = AD^2 + \Big(\dfrac{9}{2}\Big)^2 \\[1em] \Rightarrow 81 = AD^2 + \Big(\dfrac{81}{4}\Big) \\[1em] \Rightarrow AD^2 = 81 - \dfrac{81}{4} \\[1em] \Rightarrow AD^2 = \dfrac{324 - 81}{4}\\[1em] \Rightarrow AD^2 = \dfrac{243}{4} \\[1em] \Rightarrow AD = \sqrt{\dfrac{243}{4}} = \dfrac{9\sqrt{3}}{2}.$

We know that,

In an equilateral triangle, the centroid and circumcentre coincide.

Since AD is a median, and centroid divides median in 2:1 ratio,

AO : OD = 2 : 1

∴ Radius (AO) = $\dfrac{2}{3} \times AD$

= $\dfrac{2}{3} \times \dfrac{9\sqrt{3}}{2} = 3\sqrt3$ cm.

Hence, the radius of the circle = $3\sqrt{3}$ cm.