17N.2.AHL.TZ0.H_3

pestleMathematicsAAHLPaper 217N· sl-3-4-circle-radians-arcs-sectorssource ↗

This diagram shows a metallic pendant made out of four equal sectors of a larger circle of radius ${\text{OB}} = 9{\text{ cm}}$ and four equal sectors of a smaller circle of radius ${\text{OA}} = 3{\text{ cm}}$ .
The angle ${\text{BOC}} =$ 20°.

N17/5/MATHL/HP2/ENG/TZ0/03

Find the area of the pendant.

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

area = (four sector areas radius 9) + (four sector areas radius 3) (M1)

$= 4\left( {\frac{1}{2}{9^2}\frac{\pi }{9}} \right) + 4\left( {\frac{1}{2}{3^2}\frac{{7\pi }}{{18}}} \right)$ (A1)(A1)

$= 18\pi + 7\pi$

$= 25\pi {\text{ }}( = 78.5{\text{ c}}{{\text{m}}^2})$ A1

METHOD 2

area =

(area of circle radius 3) + (four sector areas radius 9) – (four sector areas radius 3) (M1)

$\pi {3^2} + 4\left( {\frac{1}{2}{9^2}\frac{\pi }{9}} \right) - 4\left( {\frac{1}{2}{3^2}\frac{\pi }{9}} \right)$ (A1)(A1)

Note: Award A1 for the second term and A1 for the third term.

$= 9\pi + 18\pi - 2\pi$

$= 25\pi {\text{ }}( = {\text{ }}78.5{\text{ c}}{{\text{m}}^2})$ A1

Note: Accept working in degrees.

[4 marks]

Examiners’ report

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