| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2010 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Triangle and sector combined - area/perimeter with given values |
| Difficulty | Moderate -0.3 This is a straightforward application of standard sector formulas (area = ½r²θ, arc length = rθ) followed by basic triangle calculations using cosine rule. Part (a) involves direct formula substitution with given values, while part (b) requires finding AP using the cosine rule then adding lengths. All techniques are routine C2 content with no novel problem-solving required, making it slightly easier than average. |
| 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 \(= \frac{1}{2}r^2\theta\) | M1 | Stated or explicitly used |
| \(= \frac{1}{2} \times 15^2 \times 1.2 = 135\) (cm²) | A1 | AG Must see some substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Arc \(= r\theta\) | M1 | PI |
| \(= 18\) (cm) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(PB = 5\) (cm) | B1 | Accept even if only on diagram or within expression for perimeter |
| \(AP^2 = 15^2 + 10^2 - 2\times15\times10\cos1.2\) | M1 | RHS of cosine rule used |
| \(= 325 - 300\cos1.2 = 216.2926\ldots\) | m1 | Correct order of evaluation |
| \(AP = 14.7(068\ldots)\) | A1 | PI eg within expression for perimeter |
| Perimeter \(= 5 + 18 + 14.7\ldots = 37.7\) (cm) | A1 | 3sf or better |
## Question 1:
**Part (a)(i)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Area of sector $= \frac{1}{2}r^2\theta$ | M1 | Stated or explicitly used |
| $= \frac{1}{2} \times 15^2 \times 1.2 = 135$ (cm²) | A1 | AG Must see some substitution |
**Part (a)(ii)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Arc $= r\theta$ | M1 | PI |
| $= 18$ (cm) | A1 | |
**Part (b)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $PB = 5$ (cm) | B1 | Accept even if only on diagram or within expression for perimeter |
| $AP^2 = 15^2 + 10^2 - 2\times15\times10\cos1.2$ | M1 | RHS of cosine rule used |
| $= 325 - 300\cos1.2 = 216.2926\ldots$ | m1 | Correct order of evaluation |
| $AP = 14.7(068\ldots)$ | A1 | PI eg within expression for perimeter |
| Perimeter $= 5 + 18 + 14.7\ldots = 37.7$ (cm) | A1 | 3sf or better |
---1 The diagram shows a sector $O A B$ of a circle with centre $O$.\\
\includegraphics[max width=\textwidth, alt={}, center]{961ff4d6-b62a-4fab-8204-8a33a969d343-2_444_373_541_804}
The radius of the circle is 15 cm and angle $A O B = 1.2$ radians.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that the area of the sector is $135 \mathrm {~cm} ^ { 2 }$.
\item Calculate the length of the arc $A B$.
\end{enumerate}\item The point $P$ lies on the radius $O B$ such that $O P = 10 \mathrm {~cm}$, as shown in the diagram below.\\
\includegraphics[max width=\textwidth, alt={}, center]{961ff4d6-b62a-4fab-8204-8a33a969d343-2_449_378_1436_799}
Calculate the perimeter of the shaded region bounded by $A P , P B$ and the arc $A B$, giving your answer to three significant figures.\\
(5 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2010 Q1 [9]}}