| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2005 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Sector area calculation |
| Difficulty | Moderate -0.8 This is a straightforward application of the sector area formula (A = ½r²θ) requiring simple algebraic rearrangement to find θ, followed by routine calculation of perimeter (2r + rθ). Both parts involve direct formula recall with minimal problem-solving, making it easier than average. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(2.4, 2\frac{2}{5}, \frac{12}{5}\) | B3 | M1 for \(30 = \frac{1}{2} × 25 × \theta\) o.e. M1 for \(\theta = (2 × 30) / 5^2\) |
| (ii) \(22\) | P2 | M1 for (arc \() = 5 ×\) their \(2.4\) |
| 5 marks |
(i) $2.4, 2\frac{2}{5}, \frac{12}{5}$ | B3 | M1 for $30 = \frac{1}{2} × 25 × \theta$ o.e. M1 for $\theta = (2 × 30) / 5^2$
(ii) $22$ | P2 | M1 for (arc $) = 5 ×$ their $2.4$
| | 5 marks |7
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{72b4624f-e716-4a37-96f3-01b46e0bd0fd-4_305_897_310_795}
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\caption{Fig. 7}
\end{center}
\end{figure}
Fig. 7 shows a sector of a circle of radius 5 cm which has angle $\theta$ radians. The sector has area $30 \mathrm {~cm} ^ { 2 }$.\\
(i) Find $\theta$.\\
(ii) Hence find the perimeter of the sector.
\hfill \mbox{\textit{OCR MEI C2 2005 Q7 [5]}}