| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2012 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Sector with attached triangle |
| Difficulty | Standard +0.3 This is a standard C2 sector question with straightforward applications of arc length (s=rθ) and sector area (A=½r²θ) formulas in parts (a)-(b). Part (c) requires solving an isosceles triangle using cosine rule, which is given as 'show that'. Parts (d)-(e) combine the sector with triangle geometry but follow routine methods. Slightly above average due to the multi-part nature and combining topics, but all techniques are standard C2 content with no novel insight required. |
| 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 |
| \(r\theta = 6 \times 0.95 = 5.7\) (cm) | M1, A1 | \(\theta\) must be in radians; A1 does not need units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{2}r^2\theta = \frac{1}{2} \times 6^2 \times 0.95 = 17.1\) (cm²) | M1, A1 | \(\theta\) must be in radians; A1 does not need units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Let \(AD = x\), then \(\frac{x}{\sin 0.95} = \frac{6}{\sin 1.24}\), so \(x = 5.16\) | M1, A1 | Complete correct trig method; accept answers rounding to 5.16 |
| OR: \(x = 3/\cos 0.95\) or \(x = 3/\sin 0.62\), so \(x = 5.16\) | If 5.16 assumed and verified: M1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Perimeter \(= 5.7 + 5.16 + 6 - 5.16 = 11.7\), or \(6 +\) their \(5.7\) | M1A1ft | A1 can follow wrong answer from (c) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Area of triangle \(ABD = \frac{1}{2} \times 6 \times 5.16 \times \sin 0.95 = 12.6\) | M1, A1 | Accept awrt 12.6 |
| Area of \(R = 17.1 - 12.6 = 4.5\) | M1, A1 | M1: area of sector minus area of triangle \(ABD\) |
## Question 7:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $r\theta = 6 \times 0.95 = 5.7$ (cm) | M1, A1 | $\theta$ must be in radians; A1 does not need units |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}r^2\theta = \frac{1}{2} \times 6^2 \times 0.95 = 17.1$ (cm²) | M1, A1 | $\theta$ must be in radians; A1 does not need units |
### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Let $AD = x$, then $\frac{x}{\sin 0.95} = \frac{6}{\sin 1.24}$, so $x = 5.16$ | M1, A1 | Complete correct trig method; accept answers rounding to 5.16 |
| OR: $x = 3/\cos 0.95$ or $x = 3/\sin 0.62$, so $x = 5.16$ | | If 5.16 assumed and verified: M1A0 |
### Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Perimeter $= 5.7 + 5.16 + 6 - 5.16 = 11.7$, or $6 +$ their $5.7$ | M1A1ft | A1 can follow wrong answer from (c) |
### Part (e)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Area of triangle $ABD = \frac{1}{2} \times 6 \times 5.16 \times \sin 0.95 = 12.6$ | M1, A1 | Accept awrt 12.6 |
| Area of $R = 17.1 - 12.6 = 4.5$ | M1, A1 | M1: area of sector minus area of triangle $ABD$ |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{42116a65-60ec-4dff-a05e-bab529939e1e-09_408_435_262_756}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows $A B C$, a sector of a circle of radius 6 cm with centre $A$. Given that the size of angle $B A C$ is 0.95 radians, find
\begin{enumerate}[label=(\alph*)]
\item the length of the $\operatorname { arc } B C$,
\item the area of the sector $A B C$.
The point $D$ lies on the line $A C$ and is such that $A D = B D$. The region $R$, shown shaded in Figure 2, is bounded by the lines $C D , D B$ and the $\operatorname { arc } B C$.
\item Show that the length of $A D$ is 5.16 cm to 3 significant figures.
Find
\item the perimeter of $R$,
\item the area of $R$, giving your answer to 2 significant figures.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2012 Q7 [12]}}