| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Circular arc problems |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing basic coordinate geometry (distance formula), arc length/sector area formulas, and volume of a prism. All steps are routine applications of standard formulas with clear guidance ('show that' parts give the answers). Slightly easier than average due to scaffolding and computational nature. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.10c Magnitude and direction: of vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(BM = \sqrt{7^2 + 24^2} = 25\) | B1 | (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\tan\alpha = \dfrac{7}{24}\) or equiv. and \(\angle BMC = 2\alpha\), or cosine rule | M1 A1 | |
| \(\angle BMC = 0.568\) radians | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\triangle ABM\): \(\frac{1}{2}(14 \times 24) = 168\ \text{mm}^2\) (or other appropriate \(\triangle\)) | B1 | |
| Sector: \(\frac{1}{2}(25^2 \times 0.568)\) | M1 A1 | |
| Total: \(168 + 168 + 177.5 = 513\ \text{mm}^2\) (or 514, or 510) | M1 A1 | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Volume \(= 513 \times 85\ \text{mm}^3\) (M requires unit conversion) | M1 | |
| \(= 44\ \text{cm}^3\) | A1 | (2) |
## Question 7:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $BM = \sqrt{7^2 + 24^2} = 25$ | B1 | (1) |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\tan\alpha = \dfrac{7}{24}$ or equiv. and $\angle BMC = 2\alpha$, or cosine rule | M1 A1 | |
| $\angle BMC = 0.568$ radians | A1 | (3) |
### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\triangle ABM$: $\frac{1}{2}(14 \times 24) = 168\ \text{mm}^2$ (or other appropriate $\triangle$) | B1 | |
| Sector: $\frac{1}{2}(25^2 \times 0.568)$ | M1 A1 | |
| Total: $168 + 168 + 177.5 = 513\ \text{mm}^2$ (or 514, or 510) | M1 A1 | (5) |
### Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Volume $= 513 \times 85\ \text{mm}^3$ (M requires unit conversion) | M1 | |
| $= 44\ \text{cm}^3$ | A1 | (2) |
---7.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\includegraphics[alt={},max width=\textwidth]{ba5cb933-dedd-4ad9-9e66-49636870b3de-3_739_1272_826_328}
\end{center}
\end{figure}
Fig. 1 shows the cross-section $A B C D$ of a chocolate bar, where $A B , C D$ and $A D$ are straight lines and $M$ is the mid-point of $A D$. The length $A D$ is 28 mm , and $B C$ is an arc of a circle with centre $M$. Taking $A$ as the origin, $B , C$ and $D$ have coordinates (7,24), (21,24) and (28,0) respectively.
\begin{enumerate}[label=(\alph*)]
\item Show that the length of $B M$ is 25 mm .
\item Show that, to 3 significant figures, $\angle B M C = 0.568$ radians.
\item Hence calculate, in $\mathrm { mm } ^ { 2 }$, the area of the cross-section of the chocolate bar.
Given that this chocolate bar has length 85 mm ,
\item calculate, to the nearest $\mathrm { cm } ^ { 3 }$, the volume of the bar.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q7 [11]}}