| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Tangent and sector - two tangents from external point |
| Difficulty | Standard +0.3 Part (a) is a direct application of arc length formula (s=rθ). Part (b)(i) uses angle sum in quadrilateral (standard geometry). Part (b)(ii) requires finding tangent length using trigonometry, then calculating area of triangle minus sector - a multi-step problem but using routine C2 techniques with no novel insight required. Slightly above average due to the geometric reasoning needed. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Arc} = r\theta = 18 \times \frac{2\pi}{3} = 12\pi\) (m) | M1, A1 (2 marks) | \(r\theta\) seen or used for arc length; \(12\pi\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\alpha = \frac{\pi}{3}\) | B1 (1 mark) | \(\frac{1}{3}\pi\) OE expression simplifying to \(\frac{1}{3}\pi\) |
| Answer | Marks | Guidance |
|---|---|---|
| Area of sector \(= \frac{1}{2}r^2\theta = \frac{1}{2} \times 18^2 \times \frac{2\pi}{3} = 108\pi\) | M1, A1 | \(\frac{1}{2}r^2\theta\) seen or used; if not exact accept 3sf or better |
| \(\tan\frac{\pi}{3} = \frac{TP}{18}\); \(TP = 18\sqrt{3} = 31.1769...\) | M1, A1 | Correct \(TP\) or \(TQ\) or \(PQ\) or \(\frac{1}{2}PQ\) or \(OT\) |
| Area of kite \(PTQO = 2 \times \frac{1}{2} \times 18 \times TP = 324\sqrt{3}\) | M1 | OE valid method for kite area |
| Area of shaded region \(= 561.(18...) - 108\pi = 222 \text{ m}^2\) to 3sf | A1 (6 marks) | If not 222, condone value from 221.7 to 222.0 inclusive |
## Question 5:
### Part (a):
| $\text{Arc} = r\theta = 18 \times \frac{2\pi}{3} = 12\pi$ (m) | M1, A1 (2 marks) | $r\theta$ seen or used for arc length; $12\pi$ |
### Part (b)(i):
| $\alpha = \frac{\pi}{3}$ | B1 (1 mark) | $\frac{1}{3}\pi$ OE expression simplifying to $\frac{1}{3}\pi$ |
### Part (b)(ii):
| Area of sector $= \frac{1}{2}r^2\theta = \frac{1}{2} \times 18^2 \times \frac{2\pi}{3} = 108\pi$ | M1, A1 | $\frac{1}{2}r^2\theta$ seen or used; if not exact accept 3sf or better |
| $\tan\frac{\pi}{3} = \frac{TP}{18}$; $TP = 18\sqrt{3} = 31.1769...$ | M1, A1 | Correct $TP$ or $TQ$ or $PQ$ or $\frac{1}{2}PQ$ or $OT$ |
| Area of kite $PTQO = 2 \times \frac{1}{2} \times 18 \times TP = 324\sqrt{3}$ | M1 | OE valid method for kite area |
| Area of shaded region $= 561.(18...) - 108\pi = 222 \text{ m}^2$ to 3sf | A1 (6 marks) | If not 222, condone value from 221.7 to 222.0 inclusive |
---5 The diagram shows a sector $O P Q$ of a circle with centre $O$.\\
\includegraphics[max width=\textwidth, alt={}, center]{a5fa3066-e330-46d0-98e3-92d438ed6f61-3_305_531_1105_758}
The radius of the circle is 18 m and the angle $P O Q$ is $\frac { 2 \pi } { 3 }$ radians.
\begin{enumerate}[label=(\alph*)]
\item Find the length of the arc $P Q$, giving your answer as a multiple of $\pi$.
\item The tangents to the circle at the points $P$ and $Q$ meet at the point $T$, and the angles $T P O$ and $T Q O$ are both right angles, as shown in the diagram below.\\
\includegraphics[max width=\textwidth, alt={}, center]{a5fa3066-e330-46d0-98e3-92d438ed6f61-3_597_529_1848_758}
\begin{enumerate}[label=(\roman*)]
\item Angle $P T Q = \alpha$ radians. Find $\alpha$ in terms of $\pi$.
\item Find the area of the shaded region bounded by the $\operatorname { arc } P Q$ and the tangents $T P$ and $T Q$, giving your answer to three significant figures.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2012 Q5 [9]}}