From a vertex of an equilateral triangle of side $2x$, a circular arc is drawn to divide the triangle into two regions, as shown in the diagram below.

Given that the areas of the two regions are equal, find the radius of the arc in terms of x.

area of triangle $= \frac{1}{2}{(2x)^2}\sin \frac{\pi }{3}$     (M1)

$= {x^2}\sqrt 3$     A1

Note: A $0.5 \times {\text{base}} \times {\text{height}}$ calculation is acceptable.

area of sector ${\text{ = }}\frac{\theta }{2}{r^2} = \frac{\pi }{6}{r^2}$     (M1)A1

area of triangle is twice the area of the sector

$\Rightarrow 2\left( {\frac{\pi }{6}{r^2}} \right) = {x^2}\sqrt 3$     M1

$\Rightarrow r = x\sqrt {\frac{{3\sqrt 3 }}{\pi }} \,\,\,\,\,$or equivalent     A1

[6 marks]

The majority of candidates obtained the correct answer. A small minority of candidates used degree measure rather than radian measure, or failed to notice that the triangle was equilateral.