| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Segment area calculation |
| Difficulty | Moderate -0.3 This is a straightforward sector/segment problem requiring standard formulas. Part (a) involves finding the radius from perimeter (20 = 2r + 8, so r = 6), then calculating chord length using cosine rule. Part (b) uses segment area = sector area - triangle area. All steps are routine applications of memorized formulas with no conceptual challenges, making it slightly easier than average. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([2r+8=20 \Rightarrow] r=6\) | B1 | |
| Angle \(AOB = \frac{8}{\text{their } 6}\) | \*M1 | Expect \(\frac{4}{3}\) OE (76.4°). M0 Assume triangle is equilateral. |
| \(AB = 2\times6\sin\text{their }\frac{2}{3}\) or \(\sqrt{6^2+6^2-2\times6^2\cos\text{their }\frac{4}{3}}\) or \(AB=\frac{6}{\sin(\frac{\pi}{2}-\text{their }\frac{2}{3})}\times\sin\text{their }\frac{4}{3}\) | DM1 | For 6 read *their* \(r\). |
| Perimeter \(=[7.42+8=]15.4\) | A1 | AWRT |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Area \(=\frac{1}{2}\times6^2\times\text{their }\frac{4}{3}-\frac{1}{2}\times6^2\times\sin\text{their }\frac{4}{3}\) or Area \(=\frac{1}{2}\times6^2\times\text{their }\frac{4}{3}-2\times\frac{1}{2}(6\sin\text{their }\frac{2}{3})(6\cos\text{their }\frac{2}{3})\) | M1 | Sector area − whole triangle area. For 6 read *their* \(r\). Sector area − 2(half triangle area). |
| \(=[24-17.49=]6.51\) | A1 | AWRT |
| 2 |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[2r+8=20 \Rightarrow] r=6$ | **B1** | |
| Angle $AOB = \frac{8}{\text{their } 6}$ | **\*M1** | Expect $\frac{4}{3}$ OE (76.4°). M0 Assume triangle is equilateral. |
| $AB = 2\times6\sin\text{their }\frac{2}{3}$ or $\sqrt{6^2+6^2-2\times6^2\cos\text{their }\frac{4}{3}}$ or $AB=\frac{6}{\sin(\frac{\pi}{2}-\text{their }\frac{2}{3})}\times\sin\text{their }\frac{4}{3}$ | **DM1** | For 6 read *their* $r$. |
| Perimeter $=[7.42+8=]15.4$ | **A1** | AWRT |
| | **4** | |
---
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Area $=\frac{1}{2}\times6^2\times\text{their }\frac{4}{3}-\frac{1}{2}\times6^2\times\sin\text{their }\frac{4}{3}$ or Area $=\frac{1}{2}\times6^2\times\text{their }\frac{4}{3}-2\times\frac{1}{2}(6\sin\text{their }\frac{2}{3})(6\cos\text{their }\frac{2}{3})$ | **M1** | Sector area − whole triangle area. For 6 read *their* $r$. Sector area − 2(half triangle area). |
| $=[24-17.49=]6.51$ | **A1** | AWRT |
| | **2** | |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{0f59214c-df46-46a6-ae9e-b9c8f104e159-06_494_542_260_799}
The diagram shows a sector $O A B$ of a circle with centre $O$. The length of the $\operatorname { arc } A B$ is 8 cm . It is given that the perimeter of the sector is 20 cm .
\begin{enumerate}[label=(\alph*)]
\item Find the perimeter of the shaded segment.
\item Find the area of the shaded segment.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2022 Q5 [6]}}