C2 June 2010 Q6
6.
Figure 1 shows the sector \(OAB\) of a circle with centre \(O\), radius 9 cm and angle 0.7 radians.
The line \(AC\) shown in Figure 1 is perpendicular to \(OA\), and \(OBC\) is a straight line.
The region \(H\) is bounded by the arc \(AB\) and the lines \(AC\) and \(CB\).
| Scheme | Marks |
|---|---|
| (a) \(r\theta = 9 \times 0.7 = 6.3\) (Also allow 6.30, or awrt 6.30) | M1 A1 |
| (2) |
Notes
(a) M: Use of \(r\theta\) (with \(\theta\) in radians), or equivalent (could be working in degrees with a correct degrees formula).
| Scheme | Marks |
|---|---|
| (b) \(\dfrac{1}{2}r^2\theta = \dfrac{1}{2} \times 81 \times 0.7 = 28.35\) (Also allow 28.3 or 28.4, or awrt 28.3 or 28.4) (Condone \(28.35^2\) written instead of \(28.35\,\text{cm}^2\)) | M1 A1 |
| (2) |
Notes
(b) M: Use of \(\dfrac{1}{2}r^2\theta\) (with \(\theta\) in radians), or equivalent (could be working in degrees with a correct degrees formula).
| Scheme | Marks |
|---|---|
| (c) \(\tan 0.7 = \dfrac{AC}{9}\) | M1 |
| \(AC = 7.58\) (Allow awrt) NOT 7.59 (see below) | A1 |
| (2) |
Notes
(c) M: Other methods must be fully correct,
e.g. \(\dfrac{AC}{\sin 0.7} = \dfrac{9}{\sin\left(\frac{\pi}{2} - 0.7\right)}\)
\((\pi - 0.7)\) instead of \(\left(\dfrac{\pi}{2} - 0.7\right)\) here is not a fully correct method.
Premature approximation (e.g. taking angle \(C\) as 0.87 radians):
This will often result in loss of A marks, e.g. \(AC = 7.59\) in (c) is A0.
| Scheme | Marks |
|---|---|
| (d) Area of triangle \(AOC = \dfrac{1}{2}(9 \times \text{their } AC)\) (or other complete method) | M1 |
| Area of \(H\) = "34.11" – "28.35" (triangle – sector) or (sector – triangle) (needs a value for each) | M1 |
| \(= 5.76\) (Allow awrt) | A1 |
| (3) | |
| 9 |
Notes
(corrected from the printed mark scheme: it prints “Area of \(R\)”; the region in this question is \(H\).)