JAMB 2000 · UME · Q27

An equilateral triangle of side $\sqrt3$ cm is inscribed in a circle. Find the radius of the circle.

Worked solution (try it first)

The height of an equilateral triangle of side $\sqrt3$ is $\sqrt3 \sin60^\circ = \sqrt3 \times \frac{\sqrt3}{2}$
$= \frac32$ cm.
The centre of the circle is where the medians meet, two-thirds of the way down each height from the vertex.
So the radius is $\frac23 \times \frac32 = 1$ cm, option C.