| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Sector perimeter calculation |
| Difficulty | Easy -1.2 This is a straightforward application of standard radian formulas (area = ½r²θ, arc length = rθ) with simple arithmetic. Part (b)(ii) requires recognizing that perimeter = 2r + rθ = r(2 + θ), leading to k=4, but this is routine algebraic manipulation rather than genuine problem-solving. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Area of sector \(= \frac{1}{2}r^2\theta = \frac{1}{2} \times 6^2 \times 0.5\) | M1 | \(\frac{1}{2}r^2\theta\) seen within (a) or used for the area |
| \(= 9 \text{ (cm}^2)\) | A1 | Condone missing/incorrect units |
| Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Arc \(= r\theta = 6 \times 0.5 = 3\) (cm) | M1, A1 | \(r\theta\) seen within (b) or used for arc length. Condone missing/incorrect units |
| Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Perimeter of sector \(= 6+6+\text{arc length} = 15\) (cm) \((= 5\times3)\) | M1 | PI by value of \(12 + \) c's (b)(i) answer |
| Perimeter of sector \(= 5\times(\text{length of})\) arc | A1 | Completion, including concluding statement |
| Total: 2 marks |
# Question 2:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Area of sector $= \frac{1}{2}r^2\theta = \frac{1}{2} \times 6^2 \times 0.5$ | M1 | $\frac{1}{2}r^2\theta$ seen within (a) or used for the area |
| $= 9 \text{ (cm}^2)$ | A1 | Condone missing/incorrect units |
| **Total: 2 marks** | | |
## Part (b)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Arc $= r\theta = 6 \times 0.5 = 3$ (cm) | M1, A1 | $r\theta$ seen within (b) or used for arc length. Condone missing/incorrect units |
| **Total: 2 marks** | | |
## Part (b)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Perimeter of sector $= 6+6+\text{arc length} = 15$ (cm) $(= 5\times3)$ | M1 | PI by value of $12 + $ c's (b)(i) answer |
| Perimeter of sector $= 5\times(\text{length of})$ arc | A1 | Completion, including concluding statement |
| **Total: 2 marks** | | |
---2 The diagram shows a sector $O A B$ of a circle with centre $O$.\\
\includegraphics[max width=\textwidth, alt={}, center]{258f0400-6e3b-406c-9f86-acc9fff4e094-2_440_392_1500_826}
The radius of the circle is 6 cm and the angle $A O B = 0.5$ radians.
\begin{enumerate}[label=(\alph*)]
\item Find the area of the sector $O A B$.
\item \begin{enumerate}[label=(\roman*)]
\item Find the length of the arc $A B$.
\item Hence show that the perimeter of the sector $O A B = k \times$ the length of the $\operatorname { arc } A B$ where $k$ is an integer.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2011 Q2 [6]}}