18M.2.AHL.TZ2.H_4

pestleMathematicsAAHLPaper 218M· sl-3-4-circle-radians-arcs-sectorssource ↗

Consider the following diagram.

The sides of the equilateral triangle ABC have lengths 1 m. The midpoint of [AB] is denoted by P. The circular arc AB has centre, M, the midpoint of [CP].

Find AM.

[3]

a.i.

Find ${\text{A}}\mathop {\text{M}}\limits^ \wedge {\text{P}}$ in radians.

[2]

a.ii.

Find the area of the shaded region.

[3]

Markscheme / solution

METHOD 1

PC $= \frac{{\sqrt 3 }}{2}$ or 0.8660 (M1)

PM $= \frac{1}{2}$ PC $= \frac{{\sqrt 3 }}{4}$ or 0.4330 (A1)

AM $= \sqrt {\frac{1}{4} + \frac{3}{{16}}}$

$= \frac{{\sqrt 7 }}{4}$ or 0.661 (m) A1

METHOD 2

using the cosine rule

AM² $= {1^2} + {\left( {\frac{{\sqrt 3 }}{4}} \right)^2} - 2 \times \frac{{\sqrt 3 }}{4} \times {\text{cos}}\left( {30^\circ } \right)$ M1A1

AM $= \frac{{\sqrt 7 }}{4}$ or 0.661 (m) A1

[3 marks]

a.i.

tan ( ${\text{A}}\mathop {\text{M}}\limits^ \wedge {\text{P}}$ ) $= \frac{2}{{\sqrt 3 }}$ or equivalent (M1)

= 0.857 A1

[2 marks]

a.ii.

EITHER

$\frac{1}{2}{\text{A}}{{\text{M}}^2}\left( {2\,{\text{A}}\mathop {\text{M}}\limits^ \wedge {\text{P}} - {\text{sin}}\left( {2\,{\text{A}}\mathop {\text{M}}\limits^ \wedge {\text{P}}} \right)} \right)$ (M1)A1

$\frac{1}{2}{\text{A}}{{\text{M}}^2} \times 2\,{\text{A}}\mathop {\text{M}}\limits^ \wedge {\text{P}} - = \frac{{\sqrt 3 }}{8}$ (M1)A1

= 0.158(m²) A1

Note: Award M1 for attempting to calculate area of a sector minus area of a triangle.

[3 marks]

Examiners’ report

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a.i.

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a.ii.

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