19N.2.AHL.TZ0.H_4

pestleMathematicsAAHLPaper 219N· sl-3-4-circle-radians-arcs-sectorssource ↗

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]

the area.

[5]

Markscheme / solution

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

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

[2 marks]

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

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

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

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

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

[5 marks]

Examiners’ report

[N/A]