| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| Session | January |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Optimization with sectors |
| Difficulty | Standard +0.3 This is a straightforward sector area problem requiring students to set up an equation where half the original sector area equals a smaller sector area, then solve for the relationship between a and r. It involves standard formulas (sector area = ½r²θ) and basic algebraic manipulation including square roots, but no novel insight or complex multi-step reasoning beyond what's typical for C2 level. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| area sector \(= \frac{1}{2} \times r^2 \times \frac{\pi}{6} \left[= \frac{\pi r^2}{12}\right]\) | M1 | soi |
| area triangle \(= \frac{1}{2} \times a^2 \times \sin\frac{\pi}{6} \left[= \frac{a^2}{4}\right]\) | M1 | soi |
| \(\frac{1}{2}a^2 \times \frac{1}{2} = \frac{1}{2} \times r^2 \times \frac{\pi}{6} \times \frac{1}{2}\) | M1 | soi |
| \(\frac{a^2}{4} = \frac{\pi r^2}{24}\) o.e. and completion to given answer | A1 |
**Question 9:**
| Answer | Mark | Guidance |
|--------|------|----------|
| area sector $= \frac{1}{2} \times r^2 \times \frac{\pi}{6} \left[= \frac{\pi r^2}{12}\right]$ | **M1** | soi | |
| area triangle $= \frac{1}{2} \times a^2 \times \sin\frac{\pi}{6} \left[= \frac{a^2}{4}\right]$ | **M1** | soi | allow sin30 |
| $\frac{1}{2}a^2 \times \frac{1}{2} = \frac{1}{2} \times r^2 \times \frac{\pi}{6} \times \frac{1}{2}$ | **M1** | soi | no follow through marks available |
| $\frac{a^2}{4} = \frac{\pi r^2}{24}$ o.e. and completion to given answer | **A1** | | at least one correct intermediate step required, and no wrong working to obtain given answer |
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## Section B
---9 Charles has a slice of cake; its cross-section is a sector of a circle, as shown in Fig. 9. The radius is $r \mathrm {~cm}$ and the sector angle is $\frac { \pi } { 6 }$ radians.
He wants to give half of the slice to Jan. He makes a cut across the sector as shown.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{aa8688bb-3608-45ab-85b7-bf84889dd189-3_420_657_497_744}
\captionsetup{labelformat=empty}
\caption{Fig. 9}
\end{center}
\end{figure}
Show that when they each have half the slice, $a = r \sqrt { \frac { \pi } { 6 } }$.
Section B (36 marks)\\
\hfill \mbox{\textit{OCR MEI C2 2011 Q9 [4]}}