| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Shaded region with arc |
| Difficulty | Standard +0.3 This is a straightforward C2 question requiring basic trigonometry (isosceles triangle properties), area of triangle formula, and sector area calculations. Part (a) is routine, and part (b) is scaffolded with the answer given, requiring only standard sector area subtraction from the triangle area. Slightly above trivial due to the multi-step nature and sector calculations, but well below average difficulty as it involves standard techniques with clear guidance. |
| 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 |
|---|---|---|
| Answer/Working | Marks | Notes |
| isosceles \(\therefore \angle AMB = 90°\) | B1 | |
| \(BM = 4\tan 30° = \frac{4}{\sqrt{3}}\) | M1 A1 | |
| area \(= \frac{1}{2} \times 8 \times \frac{4}{\sqrt{3}} = \frac{16}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{16}{3}\sqrt{3}\) cm² | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| area of sector \(= \frac{1}{2} \times 4^2 \times \frac{\pi}{6} = \frac{4}{3}\pi\) | B1 M1 | |
| shaded area \(= \frac{16}{3}\sqrt{3} - (2 \times \frac{4}{3}\pi)\) | M1 | |
| \(= \frac{16}{3}\sqrt{3} - \frac{8}{3}\pi = \frac{8}{3}(2\sqrt{3} - \pi)\) cm² | A1 | (9) |
## Question 6:
**(a)**
| Answer/Working | Marks | Notes |
|---|---|---|
| isosceles $\therefore \angle AMB = 90°$ | B1 | |
| $BM = 4\tan 30° = \frac{4}{\sqrt{3}}$ | M1 A1 | |
| area $= \frac{1}{2} \times 8 \times \frac{4}{\sqrt{3}} = \frac{16}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{16}{3}\sqrt{3}$ cm² | M1 A1 | |
**(b)**
| Answer/Working | Marks | Notes |
|---|---|---|
| area of sector $= \frac{1}{2} \times 4^2 \times \frac{\pi}{6} = \frac{4}{3}\pi$ | B1 M1 | |
| shaded area $= \frac{16}{3}\sqrt{3} - (2 \times \frac{4}{3}\pi)$ | M1 | |
| $= \frac{16}{3}\sqrt{3} - \frac{8}{3}\pi = \frac{8}{3}(2\sqrt{3} - \pi)$ cm² | A1 | **(9)** |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{288b99b5-1198-4463-baed-f0a4bf03e485-3_335_890_246_456}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows triangle $A B C$ in which $A C = 8 \mathrm {~cm}$ and $\angle B A C = \angle B C A = 30 ^ { \circ }$.
\begin{enumerate}[label=(\alph*)]
\item Find the area of triangle $A B C$ in the form $k \sqrt { 3 }$.
The point $M$ is the mid-point of $A C$ and the points $N$ and $O$ lie on $A B$ and $B C$ such that $M N$ and $M O$ are arcs of circles with centres $A$ and $C$ respectively.
\item Show that the area of the shaded region $B N M O$ is $\frac { 8 } { 3 } ( 2 \sqrt { 3 } - \pi ) \mathrm { cm } ^ { 2 }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q6 [9]}}