| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2008 |
| Session | June |
| Marks | 12 |
| 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 Parts (a) and (b) are direct formula applications (arc length = rθ, sector area = ½r²θ) requiring only substitution. Parts (c) and (d) require finding BD using the cosine rule in triangle ABD, then combining with the arc length and using sector area minus triangle area. While multi-step, these are standard C2 techniques with no novel insight 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 |
|---|---|---|
| (a) | \(r\theta = 7 \times 0.8 = 5.6\) (cm) | M1 A1 (2) |
| (b) | \(\frac{1}{2}r^2\theta = \frac{1}{2} \times 7^2 \times 0.8 = 19.6\) (cm²) | M1 A1 (2) |
| (c) | \(BD^2 = 7^2 + (AD)^2 - (2 \times 7 \times AD \times \cos 0.8)\) | M1 |
| \(BD^2 = 7^2 + 3.5^2 - (2 \times 7 \times 3.5 \times \cos 0.8)\), \((BD = 5.21)\) | A1 | Or awrt 46° for the angle |
| Perimeter \(= DC + 5.6 + 5.21 = 14.3\) (cm) (Accept awrt) | M1 A1 (4) | 2nd M1: adding their \(DC\) to arc \(BC\) and their \(BD\) |
| (d) | \(\triangle ABD = \frac{1}{2} \times 7 \times (AD) \times \sin 0.8\) \((= 8.78\ldots)\) | M1 A1ft |
| Area \(= 19.6 - 8.78\ldots = 10.8\) (cm²) (Accept awrt) | M1 A1 (4) | 2nd M1: subtracting their \(\triangle ABD\) from their sector \(ABC\); note: using segment formula \(\frac{1}{2}r^2(\theta - \sin\theta)\) scores no marks |
## Question 7:
**(a)** | $r\theta = 7 \times 0.8 = 5.6$ (cm) | M1 A1 (2) | M1: use of $r\theta$ with $\theta$ in radians |
**(b)** | $\frac{1}{2}r^2\theta = \frac{1}{2} \times 7^2 \times 0.8 = 19.6$ (cm²) | M1 A1 (2) | M1: use of $\frac{1}{2}r^2\theta$ with $\theta$ in radians |
**(c)** | $BD^2 = 7^2 + (AD)^2 - (2 \times 7 \times AD \times \cos 0.8)$ | M1 | 1st M1: use of correct cosine rule to find $BD^2$ or $BD$ |
| $BD^2 = 7^2 + 3.5^2 - (2 \times 7 \times 3.5 \times \cos 0.8)$, $(BD = 5.21)$ | A1 | Or awrt 46° for the angle |
| Perimeter $= DC + 5.6 + 5.21 = 14.3$ (cm) (Accept awrt) | M1 A1 (4) | 2nd M1: adding their $DC$ to arc $BC$ and their $BD$ |
**(d)** | $\triangle ABD = \frac{1}{2} \times 7 \times (AD) \times \sin 0.8$ $(= 8.78\ldots)$ | M1 A1ft | Or awrt 46° for angle; ft their $AD$; 1st M1: use of correct area formula to find $\triangle ABD$ |
| Area $= 19.6 - 8.78\ldots = 10.8$ (cm²) (Accept awrt) | M1 A1 (4) | 2nd M1: subtracting their $\triangle ABD$ from their sector $ABC$; note: using segment formula $\frac{1}{2}r^2(\theta - \sin\theta)$ scores no marks |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0454f5f6-b5ee-40b1-bc6a-ff8aeb06a455-09_817_1029_205_484}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows $A B C$, a sector of a circle with centre $A$ and radius 7 cm .\\
Given that the size of $\angle B A C$ is exactly 0.8 radians, find
\begin{enumerate}[label=(\alph*)]
\item the length of the arc $B C$,
\item the area of the sector $A B C$.
The point $D$ is the mid-point of $A C$. The region $R$, shown shaded in Figure 1, is bounded by $C D , D B$ and the arc $B C$.
Find
\item the perimeter of $R$, giving your answer to 3 significant figures,
\item the area of $R$, giving your answer to 3 significant figures.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2008 Q7 [12]}}