| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Triangle and sector combined - area/perimeter with given values |
| Difficulty | Standard +0.3 This is a straightforward application of arc length and sector area formulas combined with basic triangle properties. Students need to find AC using cosine rule, calculate arc length CD, then find areas of triangle and sector. All steps are standard techniques with no novel problem-solving required, 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 |
|---|---|---|
| \(6\sin 0.9=\frac{AC}{2}\) or \(AC^2=6^2+6^2-2\times6\times6\cos1.8\) | M1 | OE Correct working in degrees is acceptable throughout. |
| \(AC=9.40\) | A1 | SOI. Accept \(9.39-9.41\), may be used but not seen for A1. |
| Angle \(CAB=\frac{1}{2}(\pi-1.8)\) | M1 | SOI. Expect \(0.6708\) (or \(0.671\)). |
| Arc \(CD=\) *their* \(9.40\times\) *their* \(0.6708\) | M1 | Expect \(6.306\) (or \(6.31\)), do not accept \(6\) for *their* \(AC\) or \(1.8\) for \(CAB\). |
| \([\text{Perimeter}=6+3.40+6.306=]\ 15.7\) | A1 | Accept \(15.69-15.72\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Sector \(ADC - \triangle ABC = \frac{1}{2} \times their\ 9.40^2 \times their\ 0.6708 - \frac{1}{2} \times 6^2 \times \sin 1.8\) | M1 M1 | Accept correct use of their answers from part (a). |
| \([29.64 - 17.53 =]\ 12.1\) | A1 | AWRT |
## Question 9(a):
$6\sin 0.9=\frac{AC}{2}$ or $AC^2=6^2+6^2-2\times6\times6\cos1.8$ | **M1** | OE Correct working in degrees is acceptable throughout.
$AC=9.40$ | **A1** | SOI. Accept $9.39-9.41$, may be used but not seen for A1.
Angle $CAB=\frac{1}{2}(\pi-1.8)$ | **M1** | SOI. Expect $0.6708$ (or $0.671$).
Arc $CD=$ *their* $9.40\times$ *their* $0.6708$ | **M1** | Expect $6.306$ (or $6.31$), do not accept $6$ for *their* $AC$ or $1.8$ for $CAB$.
$[\text{Perimeter}=6+3.40+6.306=]\ 15.7$ | **A1** | Accept $15.69-15.72$
**5 total**
## Question 9(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Sector $ADC - \triangle ABC = \frac{1}{2} \times their\ 9.40^2 \times their\ 0.6708 - \frac{1}{2} \times 6^2 \times \sin 1.8$ | M1 M1 | Accept correct use of their answers from part (a). |
| $[29.64 - 17.53 =]\ 12.1$ | A1 | AWRT |
---9
The diagram shows triangle $A B C$ with $A B = B C = 6 \mathrm {~cm}$ and angle $A B C = 1.8$ radians. The arc $C D$ is part of a circle with centre $A$ and $A B D$ is a straight line.
\begin{enumerate}[label=(\alph*)]
\item Find the perimeter of the shaded region.
\item Find the area of the shaded region.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2022 Q9 [8]}}