| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Sector and arc length |
| Difficulty | Moderate -0.5 This is a straightforward application of standard sector formulas (arc length s=rθ, area A=½r²θ) requiring students to solve simultaneous equations to find r and θ, then calculate triangle area by subtraction. The arithmetic is clean and the method is direct, making it slightly easier than average but still requiring multiple steps and formula manipulation. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{\frac{1}{2}r^2\theta}{r\theta} = \frac{76.8}{9.6}\) or \(\frac{1}{2}\left(\frac{9.6^2}{\theta^2}\right)\theta = 76.8\) | M1 | Eliminate \(\theta\) or \(r\) using correct formulae SOI |
| \(r = 16\) | A1 | |
| \(\theta = 0.6\) | A1 | Accept \(34.4°\) |
| \(\triangle OAB = \frac{1}{2} \times \text{their } 16^2 \times \sin \text{their } 0.6\) | M1 | Allow Segment \(= 76.8 - \frac{1}{2} \times \text{their } 16^2 \times \sin \text{their } 0.6\). Expect \(72.27\) |
| \([\text{Area} = 76.8 - 72.27 =] \ 4.53\) | A1 | AWRT |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(AB = 2 \times 16 \times \sin 0.3\) OR \(AB^2 = 16^2 + 16^2 - 2 \times 16^2 \cos 0.6\) | M1 | Any valid method with their \(r\), \(\theta\). Expect \(AB = 9.46\) |
| Perimeter \(= 9.6 + 9.46 = 19.1\) | A1 | AWRT |
## Question 6:
**Part (a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{\frac{1}{2}r^2\theta}{r\theta} = \frac{76.8}{9.6}$ or $\frac{1}{2}\left(\frac{9.6^2}{\theta^2}\right)\theta = 76.8$ | M1 | Eliminate $\theta$ or $r$ using correct formulae SOI |
| $r = 16$ | A1 | |
| $\theta = 0.6$ | A1 | Accept $34.4°$ |
| $\triangle OAB = \frac{1}{2} \times \text{their } 16^2 \times \sin \text{their } 0.6$ | M1 | Allow Segment $= 76.8 - \frac{1}{2} \times \text{their } 16^2 \times \sin \text{their } 0.6$. Expect $72.27$ |
| $[\text{Area} = 76.8 - 72.27 =] \ 4.53$ | A1 | AWRT |
**Part (b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $AB = 2 \times 16 \times \sin 0.3$ OR $AB^2 = 16^2 + 16^2 - 2 \times 16^2 \cos 0.6$ | M1 | Any valid method with their $r$, $\theta$. Expect $AB = 9.46$ |
| Perimeter $= 9.6 + 9.46 = 19.1$ | A1 | AWRT |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{51bd3ba6-e1d1-4c07-81cd-d99dd77f9306-07_389_552_267_799}
The diagram shows a sector $O A B$ of a circle with centre $O$ and radius $r \mathrm {~cm}$. Angle $A O B = \theta$ radians. It is given that the length of the $\operatorname { arc } A B$ is 9.6 cm and that the area of the sector $O A B$ is $76.8 \mathrm {~cm} ^ { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the area of the shaded region.
\item Find the perimeter of the shaded region.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2023 Q6 [7]}}