| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Segment area calculation |
| Difficulty | Moderate -0.8 This is a straightforward application of standard formulas for sector area (½r²θ) and triangle area (½r²sinθ) with all values given directly. It requires only recall of formulas and basic substitution with no problem-solving or geometric insight needed, 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) \(A = \frac{1}{2}r^2\theta = 18 \times 1.8 = 32.4\) | M1 A1 | 2 marks |
| (ii) Area triangle \(= \frac{1}{2}r^2\sin\theta = 18 \times \sin 1.8 = 17.53\) \(\Rightarrow\) Area segment \(= 32.4 - 17.53 = 14.87\) | M1 A1 A1 | 3 marks |
**(i)** $A = \frac{1}{2}r^2\theta = 18 \times 1.8 = 32.4$ | M1 A1 | 2 marks
**(ii)** Area triangle $= \frac{1}{2}r^2\sin\theta = 18 \times \sin 1.8 = 17.53$ $\Rightarrow$ Area segment $= 32.4 - 17.53 = 14.87$ | M1 A1 A1 | 3 marks8 Fig. 8 shows a sector of a circle with centre O and radius 6 cm and a chord AB which subtends an angle of 1.8 radians at O .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c55a5f04-3573-4f36-a12c-3755bdd4a45b-4_341_485_310_771}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}
(i) Calculate the area of the sector OAXB .\\
(ii) Calculate the area of the triangle OAB and hence find the area of the shaded segment AXB.
\hfill \mbox{\textit{OCR MEI C2 Q8 [5]}}