Higher June 2022 Paper 1 Q21

EdexcelCurrent spec5 marksCircles & Sectors

21 The diagram shows three circles, each of radius 4 cm.

The centres of the circles are \(A\), \(B\) and \(C\) such that \(ABC\) is a straight line and \(AB = BC = 4\) cm.

Work out the total area of the two shaded regions.
Give your answer in terms of \(\pi\) (5)

Answer	Mark	Mark scheme
\(16\sqrt{3} - \dfrac{16\pi}{3}\)	P1	for identifying an angle of 60 or 120
	P1	for process to find the area of a sector of angle 60 or 120 eg \(\pi 4^2 \times \dfrac{60}{360}\ \left(= \dfrac{8\pi}{3}\right)\) or \(\pi 4^2 \times \dfrac{120}{360}\ \left(= \dfrac{16\pi}{3}\right)\)
	P1	for process to find the area of an equilateral triangle eg \(\dfrac{1}{2} \times 4 \times 4 \times \sin 60\ (= 4\sqrt{3})\) or \(\dfrac{4 \times \sqrt{4^2 - 2^2}}{2}\ (= 2\sqrt{12} \text{ or } 4\sqrt{3})\) or the area of an isosceles triangle or area of a right-angled triangle eg \(\dfrac{1}{2} \times 4 \times 4 \times \sin 120\ (= 4\sqrt{3})\) or \(\dfrac{2 \times \sqrt{4^2 - 2^2}}{2}\ (= \sqrt{12} \text{ or } 2\sqrt{3})\)
	P1	for using area of sector – area of triangle to find area of a segment eg \(\pi 4^2 \times \dfrac{60}{360} - \dfrac{1}{2} \times 4 \times 4 \times \sin 60\ \left(= \dfrac{8\pi}{3} - 4\sqrt{3}\right)\) or \(\pi 4^2 \times \dfrac{120}{360} - \dfrac{1}{2} \times 4 \times 4 \times \sin 120\ \left(= \dfrac{16\pi}{3} - 4\sqrt{3}\right)\)
	A1	for \(16\pi - 4\left(\dfrac{16\pi}{6} - 4\sqrt{3} + \dfrac{16\pi}{6}\right)\) or \(16\sqrt{3} - \dfrac{16\pi}{3}\) oe