| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Area of region bounded by circle and line |
| Difficulty | Standard +0.3 This is a standard circle geometry problem requiring identification of angles (120° = 2π/3), arc length calculation, and area using sectors minus triangles. While it involves multiple steps and radians, the techniques are routine for P1 level with no novel geometric insight required. Slightly easier than average due to the structured parts guiding students through the solution. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(k = \frac{2}{3}\) | B1 | Allow \(\text{ACB} = \frac{2\pi}{3}\) |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Perimeter of shaded area \(= 2\pi r\) | B1 | |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Major sector \(OAB = \frac{1}{2}r^2 \times \frac{4\pi}{3}\) | *M1 | Expect \(\frac{2}{3}\pi r^2\). Finds area of any relevant sector or triangle. Can be embedded in segment formula |
| One or both segments \(= [2]\times\left(\frac{1}{2}r^2 \times \frac{\pi}{3} - \frac{1}{2}r^2\sin\frac{\pi}{3}\right)\) | *M1 | |
| \(= [2]\left(r^2\frac{\pi}{6} - r^2\frac{\sqrt{3}}{4}\right)\) | A1 | |
| Shaded area \(= \frac{2}{3}\pi r^2 - 2\left(\frac{1}{6}\pi r^2 - \frac{r^2\sqrt{3}}{4}\right)\) | DM1 | |
| \(= \frac{\pi r^2}{3} + \frac{r^2\sqrt{3}}{2}\) | A1 | |
| Total: 5 |
## Question 6:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $k = \frac{2}{3}$ | B1 | Allow $\text{ACB} = \frac{2\pi}{3}$ |
| **Total: 1** | | |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Perimeter of shaded area $= 2\pi r$ | B1 | |
| **Total: 1** | | |
### Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Major sector $OAB = \frac{1}{2}r^2 \times \frac{4\pi}{3}$ | *M1 | Expect $\frac{2}{3}\pi r^2$. Finds area of any relevant sector or triangle. Can be embedded in segment formula |
| One or both segments $= [2]\times\left(\frac{1}{2}r^2 \times \frac{\pi}{3} - \frac{1}{2}r^2\sin\frac{\pi}{3}\right)$ | *M1 | |
| $= [2]\left(r^2\frac{\pi}{6} - r^2\frac{\sqrt{3}}{4}\right)$ | A1 | |
| Shaded area $= \frac{2}{3}\pi r^2 - 2\left(\frac{1}{6}\pi r^2 - \frac{r^2\sqrt{3}}{4}\right)$ | DM1 | |
| $= \frac{\pi r^2}{3} + \frac{r^2\sqrt{3}}{2}$ | A1 | |
| **Total: 5** | | |6\\
\includegraphics[max width=\textwidth, alt={}, center]{5e3e5418-7976-4232-8550-1da6420a3fcb-08_534_506_255_815}
The diagram shows a motif formed by the major arc $A B$ of a circle with radius $r$ and centre $O$, and the minor arc $A O B$ of a circle, also with radius $r$ but with centre $C$. The point $C$ lies on the circle with centre $O$.
\begin{enumerate}[label=(\alph*)]
\item Given that angle $A C B = k \pi$ radians, state the value of the fraction $k$.
\item State the perimeter of the shaded motif in terms of $\pi$ and $r$.
\item Find the area of the shaded motif, giving your answer in terms of $\pi , r$ and $\sqrt { 3 }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2023 Q6 [7]}}