C2 June 2011 Q5
5.
The shape shown in Figure 1 is a pattern for a pendant. It consists of a sector \(OAB\) of a circle centre \(O\), of radius 6 cm, and angle \(AOB = \dfrac{\pi}{3}\). The circle \(C\), inside the sector, touches the two straight edges, \(OA\) and \(OB\), and the arc \(AB\) as shown.
Find
The region outside the circle \(C\) and inside the sector \(OAB\) is shown shaded in Figure 1.
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{2}r^2\theta = \dfrac{1}{2}(6)^2\left(\dfrac{\pi}{3}\right) = 6\pi\) or 18.85 or awrt 18.8 (cm)2 Using \(\tfrac{1}{2}r^2\theta\) (See notes) | M1 |
| \(6\pi\) or 18.85 or awrt 18.8 | A1 |
| [2] |
Notes
M1: Needs \(\theta\) in radians for this formula.
Candidate could convert to degrees and use the degrees formula.
A1: Does not need units. Answer should be either \(6\pi\) or 18.85 or awrt 18.8
Correct answer with no working is M1A1.
This M1A1 can only be awarded in part (a).
| Scheme | Marks |
|---|---|
| \(\sin\left(\dfrac{\pi}{6}\right) = \dfrac{r}{6 - r}\) \(\sin\left(\dfrac{\pi}{6}\right)\) or \(\sin 30^\circ = \dfrac{r}{6 - r}\) | M1 |
| \(\dfrac{1}{2} = \dfrac{r}{6 - r}\) Replaces sin by numeric value | dM1 |
| \(6 - r = 2r \Rightarrow r = 2\) \(r = 2\) | A1 cso |
| [3] |
Notes
M1: Also allow \(\cos\left(\dfrac{\pi}{3}\right)\) or \(\cos 60^\circ = \dfrac{r}{6 - r}\).
1st M1: Needs correct trigonometry method. Candidates could state \(\sin\left(\dfrac{\pi}{6}\right) = \dfrac{r}{x}\) and \(x + r = 6\) or equivalent in their working to gain this method mark.
dM1: Replaces sin by numerical value. \(0.009\ldots = \dfrac{r}{6 - r}\) from working “incorrectly” in degrees is fine here for dM1.
A1: For \(r = 2\) from correct solution only.
Alternative: 1st M1 for \(\tfrac{r}{OC} = \sin 30\) or \(\tfrac{r}{OC} = \cos 60\). 2nd M1 for \(OC = 2r\) and then A1 for \(r = 2\).
Note seeing \(OC = 2r\) is M1M1.
Special Case: If a candidate states an answer of \(r = 2\) (must be in part (b)) as a guess or from an incorrect method then award SC: M0M0B1. Such a candidate could then go on to score M1A1 in part (c).
| Scheme | Marks |
|---|---|
| Area \(= 6\pi - \pi(2)^2 = 2\pi\) or awrt 6.3 (cm)2 their area of sector \(- \pi r^2\) | M1 |
| \(2\pi\) or awrt 6.3 | A1 cao |
| [2] | |
| 7 |
Notes
M1: For “their area of sector – their area of circle”, where \(r \gt 0\) is ft from their answer to part (b). Allow the method mark if “their area of sector” \(\lt\) “their area of circle”. The candidate must show somewhere in their working that they are subtracting the correct way round, even if their answer is negative.
Some candidates in part (c) will either use their value of \(r\) from part (b) or even introduce a value of \(r\) in part (c). You can apply the scheme to award either M0A0 or M1A0 or M1A1 to these candidates.
Note: Candidates can get M1 by writing “their part (a) answer \(- \pi r^2\)”, where the radius of the circle is not substituted.
A1: cao – accept exact answer or awrt 6.3
Correct answer only with no working in (c) gets M1A1
Beware: The answer in (c) is the same as the arc length of the pendant