| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2014 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Shaded region between arcs |
| Difficulty | Standard +0.3 This is a standard sector/triangle combination question requiring basic radian formulas (arc length, sector area) and right-angled trigonometry. Part (i) involves expressing perimeter using arc lengths and the perpendicular AD = 4sin(α), while part (ii) requires subtracting areas of two sectors. The question is slightly easier than average as it's methodical with clear geometric setup and standard techniques, though it does require careful identification of the relevant lengths and areas. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Arc \(AB = 4\alpha\) | B1 | |
| Arc \(DC = (4\cos\alpha)\alpha\) | B1 | |
| \(AC\) (or \(DB\)) \(= 4 - 4\cos\alpha\) | B1 | |
| Perimeter \(= 4\alpha\cos\alpha + 4\alpha + 8 - 8\cos\alpha\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(OD = 4\cos\frac{\pi}{6} \left(= 2\sqrt{3}\right)\) | B1 | |
| Shaded area \(= \left[\frac{1}{2} \times 4^2 \times \frac{\pi}{6}\right] - \left[\frac{1}{2}\left(2\sqrt{3}\right)^2 \times \frac{\pi}{6}\right]\) | B1B1 | |
| \(\frac{\pi}{3}\) | B1 | Or \(k = \frac{1}{3}\) |
## Question 8:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Arc $AB = 4\alpha$ | **B1** | |
| Arc $DC = (4\cos\alpha)\alpha$ | **B1** | |
| $AC$ (or $DB$) $= 4 - 4\cos\alpha$ | **B1** | |
| Perimeter $= 4\alpha\cos\alpha + 4\alpha + 8 - 8\cos\alpha$ | **B1** | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $OD = 4\cos\frac{\pi}{6} \left(= 2\sqrt{3}\right)$ | **B1** | |
| Shaded area $= \left[\frac{1}{2} \times 4^2 \times \frac{\pi}{6}\right] - \left[\frac{1}{2}\left(2\sqrt{3}\right)^2 \times \frac{\pi}{6}\right]$ | **B1B1** | |
| $\frac{\pi}{3}$ | **B1** | Or $k = \frac{1}{3}$ |
---8\\
\includegraphics[max width=\textwidth, alt={}, center]{77543862-ed95-42bf-b788-a9a43f039a89-3_408_686_264_731}
In the diagram, $A B$ is an arc of a circle with centre $O$ and radius 4 cm . Angle $A O B$ is $\alpha$ radians. The point $D$ on $O B$ is such that $A D$ is perpendicular to $O B$. The arc $D C$, with centre $O$, meets $O A$ at $C$.\\
(i) Find an expression in terms of $\alpha$ for the perimeter of the shaded region $A B D C$.\\
(ii) For the case where $\alpha = \frac { 1 } { 6 } \pi$, find the area of the shaded region $A B D C$, giving your answer in the form $k \pi$, where $k$ is a constant to be determined.
\hfill \mbox{\textit{CAIE P1 2014 Q8 [8]}}