| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 5 |
| 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 application of sector and circle area formulas with basic algebraic manipulation. Part (i) requires setting up an equation (semicircle area = 2 × sector area) and solving for α, while part (ii) involves adding arc lengths. The question is slightly easier than average as it's a standard textbook-style problem with clear steps and no novel insight required. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \((OAB) = \frac{1}{2} \times 8^2 \times \alpha\), \((OAC) = \frac{1}{2} \times \pi \times 4^2\) | B1B1 | Accept 25.1 (for \(OAC\)) |
| \(\alpha = \frac{\pi}{8}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) \(8 + 8 \times\) their \(\alpha + \frac{1}{2} \times 8 \times \pi\) | B1√ | 23.7 gets B1B0 SC B1 for e.g. \(5\pi\) (omitted \(OB\)) |
| \(8 + 5\pi\) | B1 |
(i) $(OAB) = \frac{1}{2} \times 8^2 \times \alpha$, $(OAC) = \frac{1}{2} \times \pi \times 4^2$ | B1B1 | Accept 25.1 (for $OAC$)
$\alpha = \frac{\pi}{8}$ | B1 |
[3]
(ii) $8 + 8 \times$ their $\alpha + \frac{1}{2} \times 8 \times \pi$ | B1√ | 23.7 gets B1B0 SC B1 for e.g. $5\pi$ (omitted $OB$)
$8 + 5\pi$ | B1 |
[2]3\\
\includegraphics[max width=\textwidth, alt={}, center]{d0074ac8-42d2-49f4-a417-4a348537bccc-2_492_682_708_733}
In the diagram, $O A B$ is a sector of a circle with centre $O$ and radius 8 cm . Angle $B O A$ is $\alpha$ radians. $O A C$ is a semicircle with diameter $O A$. The area of the semicircle $O A C$ is twice the area of the sector $O A B$.\\
(i) Find $\alpha$ in terms of $\pi$.\\
(ii) Find the perimeter of the complete figure in terms of $\pi$.
\hfill \mbox{\textit{CAIE P1 2013 Q3 [5]}}