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Date	November 2019	Marks available	2	Reference code	19N.2.AHL.TZ0.H_4
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Calculate	Question number	H_4	Adapted from	N/A

Question

The following shape consists of three arcs of a circle, each with centre at the opposite vertex of an equilateral triangle as shown in the diagram.

For this shape, calculate

the perimeter.

[2]

a.

the area.

[5]

b.

Markscheme

each arc has length $r θ = 6 \times \frac{π}{3} = 2 π (= 6.283 \dots)$ $r\theta = 6 \times \frac{\pi }{3} = 2\pi \,\left( { = 6.283 \ldots } \right)$ (M1)

perimeter is therefore $6 π (= 18.8)$ $6\pi \,\left( { = 18.8} \right)$ (cm) A1

[2 marks]

a.

area of sector, $s$ $s$ , is $\frac{1}{2} r^{2} θ = 18 \times \frac{π}{3} = 6 π (= 18.84 \dots)$ $\frac{1}{2}{r^2}\theta = 18 \times \frac{\pi }{3} = 6\pi \,\left( { = 18.84 \ldots } \right)$ (A1)

area of triangle, $t$ $t$ , is $\frac{1}{2} \times 6 \times 3 \sqrt{3} = 9 \sqrt{3} (= 15.58 \dots)$ $\frac{1}{2} \times 6 \times 3\sqrt 3 = 9\sqrt 3 \,\left( { = 15.58 \ldots } \right)$ (M1)(A1)

Note: area of segment, $k$ $k$ , is 3.261… implies area of triangle

finding $3 s - 2 t$ $3s - 2t$ or $3 k + t$ $3k + t$ or similar

area $= 3 s - 2 t = 18 π - 18 \sqrt{3} (= 25.4)$ $= 3s - 2t = 18\pi - 18\sqrt 3 \,\left( { = 25.4} \right)$ (cm²) (M1)A1

[5 marks]

b.

Examiners report

[N/A]

a.

[N/A]

b.

Syllabus sections

Topic 3— Geometry and trigonometry » SL 3.4—Circle: radians, arcs, sectors

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Topic 3— Geometry and trigonometry