| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2007 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Segment area calculation |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question testing standard formulas (arc length s=rθ, sector area A=½r²θ) and the cosine rule. Part (a) is given as 'show that', parts (b-c) apply memorized formulas with minimal problem-solving. Slightly easier than average due to the scaffolded structure and routine application of techniques. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C)1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| 3(a) | Arc \(= r\theta\) | M1 |
| \(28 = 20\theta \Rightarrow \theta = 1.4\) | A1 | 2 marks |
| 3(b) | Area of sector \(= \frac{1}{2}r^2\theta\) | M1 |
| \(= \frac{1}{2} \times 20^2(1.4) = 280 \text{ (cm}^2\text{)}\) | A1 | 2 marks |
| 3(c)(i) | Area triangle \(= \frac{1}{2} \times 15 \times 20 \times \sin 1.4\) | M1 |
| Shaded area \(=\) Area of sector – area of triangle | M1 | |
| \(= 280 - 147.8\ldots = 132 \text{ (cm}^2\text{)} \text{ (3sf)}\) | A1ft | 3 marks |
| 3(c)(ii) | \(\{BD^2 = \}15^2 + 20^2 - 2 \times 15 \times 20\cos 1.4\) | M1 |
| \(= 225 + 400 - 101.98\ldots\) | m1 | |
| \(\Rightarrow BD = \sqrt{523.019\ldots} = 22.86\ldots = 22.9 \text{ (cm) to 3 sf}\) | A1 | 3 marks |
**3(a)** | Arc $= r\theta$ | M1 | |
| | $28 = 20\theta \Rightarrow \theta = 1.4$ | A1 | 2 marks | AG | For $r\theta$ or $20\theta$ or PI by $20 \times 1.4$ |
**3(b)** | Area of sector $= \frac{1}{2}r^2\theta$ | M1 | | $\frac{1}{2}r^2\theta$ OE seen |
| | $= \frac{1}{2} \times 20^2(1.4) = 280 \text{ (cm}^2\text{)}$ | A1 | 2 marks | Condone absent cm$^2$ |
**3(c)(i)** | Area triangle $= \frac{1}{2} \times 15 \times 20 \times \sin 1.4$ | M1 | | Use of $\frac{1}{2}ab\sin C$ OE |
| | Shaded area $=$ Area of sector – area of triangle | M1 | |
| | $= 280 - 147.8\ldots = 132 \text{ (cm}^2\text{)} \text{ (3sf)}$ | A1ft | 3 marks | Ft on [ans (b) $-$ 147.8...] to 3sf provided [...] $> 0$ |
**3(c)(ii)** | $\{BD^2 = \}15^2 + 20^2 - 2 \times 15 \times 20\cos 1.4$ | M1 | | RHS of cosine rule used |
| | $= 225 + 400 - 101.98\ldots$ | m1 | | Correct order of evaluation |
| | $\Rightarrow BD = \sqrt{523.019\ldots} = 22.86\ldots = 22.9 \text{ (cm) to 3 sf}$ | A1 | 3 marks | Condone absent cm |
---3 The diagram shows a sector $O A B$ of a circle with centre $O$ and radius 20 cm . The angle between the radii $O A$ and $O B$ is $\theta$ radians.\\
\includegraphics[max width=\textwidth, alt={}, center]{ad574bde-3bf1-45be-a454-9c723088b357-3_453_499_429_804}
The length of the $\operatorname { arc } A B$ is 28 cm .
\begin{enumerate}[label=(\alph*)]
\item Show that $\theta = 1.4$.
\item Find the area of the sector $O A B$.
\item The point $D$ lies on $O A$. The region bounded by the line $B D$, the line $D A$ and the arc $A B$ is shaded.\\
\includegraphics[max width=\textwidth, alt={}, center]{ad574bde-3bf1-45be-a454-9c723088b357-3_440_380_1372_806}
The length of $O D$ is 15 cm .
\begin{enumerate}[label=(\roman*)]
\item Find the area of the shaded region, giving your answer to three significant figures.\\
(3 marks)
\item Use the cosine rule to calculate the length of $B D$, giving your answer to three significant figures.\\
(3 marks)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2007 Q3 [10]}}