User interface language: English | Español

Date May 2018 Marks available 2 Reference code 18M.2.hl.TZ2.4
Level HL only Paper 2 Time zone TZ2
Command term Find Question number 4 Adapted from N/A

Question

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]
b.

Markscheme

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

OR

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

= 0.158(m2)      A1

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

[3 marks]

b.

Examiners report

[N/A]
a.i.
[N/A]
a.ii.
[N/A]
b.

Syllabus sections

Topic 3 - Core: Circular functions and trigonometry » 3.1 » The circle: radian measure of angles.
Show 23 related questions

View options