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.
Find \({\text{A}}\mathop {\text{M}}\limits^ \wedge {\text{P}}\) in radians.
Find the area of the shaded region.
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]
tan (\({\text{A}}\mathop {\text{M}}\limits^ \wedge {\text{P}}\)) \( = \frac{2}{{\sqrt 3 }}\) or equivalent (M1)
= 0.857 A1
[2 marks]
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]