| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Compound shape area |
| Difficulty | Standard +0.8 This is a multi-part question requiring sector area formula, geometric reasoning to find the inscribed circle radius using angle bisector properties and trigonometry, then combining these for the final area. The inscribed circle part (b) requires non-trivial geometric insight beyond standard formulas, elevating this above routine C2 questions. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{1}{2}r^2\theta = \frac{1}{2}(6)^2\left(\frac{\pi}{3}\right) = 6\pi\) or 18.85 or awrt 18.8 (cm)\(^2\) | M1 | Using \(\frac{1}{2}r^2\theta\); needs \(\theta\) in radians |
| \(6\pi\) or 18.85 or awrt 18.8 | A1 | Does not need units [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\sin\left(\frac{\pi}{6}\right) = \frac{r}{6-r}\) | M1 | \(\sin\left(\frac{\pi}{6}\right)\) or \(\sin 30° = \frac{r}{6-r}\) |
| \(\frac{1}{2} = \frac{r}{6-r}\) | dM1 | Replaces sin by numeric value |
| \(6 - r = 2r \Rightarrow r = 2\) | A1 cso | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Area \(= 6\pi - \pi(2)^2 = 2\pi\) or awrt 6.3 (cm)\(^2\) | M1 | Their area of sector \(- \pi r^2\) |
| \(2\pi\) or awrt 6.3 | A1 cao | [2] |
## Question 5:
### Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{1}{2}r^2\theta = \frac{1}{2}(6)^2\left(\frac{\pi}{3}\right) = 6\pi$ or 18.85 or awrt 18.8 (cm)$^2$ | M1 | Using $\frac{1}{2}r^2\theta$; needs $\theta$ in radians |
| $6\pi$ or 18.85 or awrt 18.8 | A1 | Does not need units **[2]** |
### Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $\sin\left(\frac{\pi}{6}\right) = \frac{r}{6-r}$ | M1 | $\sin\left(\frac{\pi}{6}\right)$ or $\sin 30° = \frac{r}{6-r}$ |
| $\frac{1}{2} = \frac{r}{6-r}$ | dM1 | Replaces sin by numeric value |
| $6 - r = 2r \Rightarrow r = 2$ | A1 cso | **[3]** |
### Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| Area $= 6\pi - \pi(2)^2 = 2\pi$ or awrt 6.3 (cm)$^2$ | M1 | Their area of sector $- \pi r^2$ |
| $2\pi$ or awrt 6.3 | A1 cao | **[2]** |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c9758792-ca4c-4837-bd7c-e695fe0c0cdf-06_426_417_260_760}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The shape shown in Figure 1 is a pattern for a pendant. It consists of a sector $O A B$ of a circle centre $O$, of radius 6 cm , and angle $A O B = \frac { \pi } { 3 }$. The circle $C$, inside the sector, touches the two straight edges, $O A$ and $O B$, and the $\operatorname { arc } A B$ as shown.
Find
\begin{enumerate}[label=(\alph*)]
\item the area of the sector $O A B$,
\item the radius of the circle $C$.
The region outside the circle $C$ and inside the sector $O A B$ is shown shaded in Figure 1.
\item Find the area of the shaded region.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2011 Q5 [7]}}