| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Session | Specimen |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area of sector/segment problems |
| Difficulty | Standard +0.3 This question requires identifying the circle's center and radius, recognizing the geometry of a circular segment, and applying the standard formula (sector area minus triangle area). While it involves multiple steps and careful geometric reasoning, the techniques are standard A-level content with no novel insights required. The 'show detailed reasoning' requirement adds slight complexity, but this remains an accessible application question slightly easier than average. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Radius \(= \sqrt{10}\) | B1 | 1.1 |
| \(\cos C = \frac{10+10-(7-1)^2}{2 \times \sqrt{10} \times \sqrt{10}}\) | M1 | 3.1a |
| \(C = 2.50\) (3sf) | A1 | 1.1 |
| Area \(= \frac{1}{2} \times (\sqrt{10})^2 \times 2.50 - \frac{1}{2} \times (\sqrt{10})^2 \times \sin 2.50\) | M1 | 3.1a |
| Area \(= 9.49\) | A1 | 1.1 |
## Question 5:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Radius $= \sqrt{10}$ | B1 | 1.1 |
| $\cos C = \frac{10+10-(7-1)^2}{2 \times \sqrt{10} \times \sqrt{10}}$ | M1 | 3.1a | Or use right angled triangle: M1 for $\cos x = \frac{3}{\sqrt{10}}$ and $\frac{1}{2}C = \frac{\pi}{2} - \cos^{-1}\left(\frac{3}{\sqrt{10}}\right)$ |
| $C = 2.50$ (3sf) | A1 | 1.1 |
| Area $= \frac{1}{2} \times (\sqrt{10})^2 \times 2.50 - \frac{1}{2} \times (\sqrt{10})^2 \times \sin 2.50$ | M1 | 3.1a |
| Area $= 9.49$ | A1 | 1.1 | **[5]** |
---5 In this question you must show detailed reasoning.
Fig. 5 shows the circle with equation $( x - 4 ) ^ { 2 } + ( y - 1 ) ^ { 2 } = 10$.\\
The points $( 1,0 )$ and $( 7,0 )$ lie on the circle. The point C is the centre of the circle.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b4e10fd2-4144-4019-bf00-070f93a2b05d-05_878_1000_685_255}
\captionsetup{labelformat=empty}
\caption{Fig. 5}
\end{center}
\end{figure}
Find the area of the part of the circle below the $x$-axis.
\hfill \mbox{\textit{OCR MEI Paper 3 Q5 [5]}}