| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2014 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Tangent and sector - two tangents from external point |
| Difficulty | Standard +0.3 This is a standard circle geometry problem requiring knowledge of tangent properties, arc length, and sector area formulas. While it involves multiple steps (finding angle at centre, using tangent properties, calculating arc and area), these are routine applications of well-practiced techniques with no novel insight required. The 6 marks total and straightforward geometric setup place it slightly above average difficulty. |
| Spec | 1.03f Circle properties: angles, chords, tangents1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| \(CB \text{ or } AB = \frac{3}{\tan\frac{\pi}{6}}\) or \(3\tan\frac{\pi}{3}\) | B1 | Allow throughout for e.g. \(2\sqrt{3}, \sqrt{27}, \sqrt{3^*}, \left(\sqrt{3}\right)^*, \frac{9}{\sqrt{3}}\) |
| Arc or \(AC = 3 \times \left[\frac{2\pi}{3} \text{ or } \frac{\pi}{3}\right]\) (\(= 2\pi\) or \(\pi\)) | B1 | After B0B0 SCB1 for 16.7 |
| Perimeter = \(6\sqrt{3} + 2\pi\) oe | B1 | Their AB in form \(k\sqrt{3}\) [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Area \(OABC\) \((2) \frac{1}{2} \times 3 \times \text{their } AB\) (\(= 9\sqrt{3}\) or \(\frac{9\sqrt{3}}{2}\)) | B1 | |
| Area \(OADC\) \(\frac{1}{2} \times 3^2 \times \left[\frac{2\pi}{3} \text{ or } \frac{\pi}{3}\right]\) (\(= 3\pi\) or \(\frac{3\pi}{2}\)) | B1 | After B0B0 SCB1 for 6.16 or 6.17. Allow \(\left(\sqrt{3}\right)^* - 3\pi\) |
| Shaded area \(9\sqrt{3} - 3\pi\) oe | B1 | [3] |
### (i)
$CB \text{ or } AB = \frac{3}{\tan\frac{\pi}{6}}$ or $3\tan\frac{\pi}{3}$ | **B1** | Allow throughout for e.g. $2\sqrt{3}, \sqrt{27}, \sqrt{3^*}, \left(\sqrt{3}\right)^*, \frac{9}{\sqrt{3}}$
Arc or $AC = 3 \times \left[\frac{2\pi}{3} \text{ or } \frac{\pi}{3}\right]$ ($= 2\pi$ or $\pi$) | **B1** | After B0B0 SCB1 for 16.7
Perimeter = $6\sqrt{3} + 2\pi$ oe | **B1** | Their AB in form $k\sqrt{3}$ [3]
### (ii)
Area $OABC$ $(2) \frac{1}{2} \times 3 \times \text{their } AB$ ($= 9\sqrt{3}$ or $\frac{9\sqrt{3}}{2}$) | **B1** |
Area $OADC$ $\frac{1}{2} \times 3^2 \times \left[\frac{2\pi}{3} \text{ or } \frac{\pi}{3}\right]$ ($= 3\pi$ or $\frac{3\pi}{2}$) | **B1** | After B0B0 SCB1 for 6.16 or 6.17. Allow $\left(\sqrt{3}\right)^* - 3\pi$
Shaded area $9\sqrt{3} - 3\pi$ oe | **B1** | [3]
---\includegraphics{figure_2}
In the diagram, $OADC$ is a sector of a circle with centre $O$ and radius 3 cm. $AB$ and $CB$ are tangents to the circle and angle $ABC = \frac{1}{4}\pi$ radians. Find, giving your answer in terms of $\sqrt{3}$ and $\pi$,
\begin{enumerate}[label=(\roman*)]
\item the perimeter of the shaded region, [3]
\item the area of the shaded region. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2014 Q2 [6]}}