| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Sector and arc length |
| Difficulty | Standard +0.3 This is a standard P1 geometry question involving arc length, sector area, and basic trigonometry. Part (a) requires using cosine rule or recognizing an equilateral triangle (routine). Parts (b) and (c) involve straightforward application of arc length and area formulas with some careful bookkeeping of what to add/subtract. The symmetry simplifies the problem, and all steps follow standard procedures without requiring novel insight. |
| Spec | 1.03f Circle properties: angles, chords, tangents1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Either Let midpoint of \(PQ\) be \(H\): \(\sin HCP = \dfrac{2}{4} \Rightarrow\) Angle \(HCP = \dfrac{\pi}{6}\) | M1 | |
| Or \(\sin PSQ = \dfrac{4}{8} \Rightarrow\) Angle \(PSQ = \dfrac{\pi}{6}\) | ||
| Or using cosine rule: angle \(PCQ = \dfrac{\pi}{3}\) | ||
| Or by inspection: triangle \(PCQ\) or \(PCT\) is equilateral so angle \(PCQ = \dfrac{\pi}{3}\) | ||
| Angle \(PCS = \pi - \dfrac{\pi}{6} - \dfrac{\pi}{6} = \dfrac{2}{3}\pi\) | A1 | AG |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Perimeter \(= 2 \times 4 \times \dfrac{2\pi}{3}\) or \(8\pi - \dfrac{8\pi}{3}\) | M1 | Length of two arcs \(PS\) and \(QR\) |
| \(+2\pi \times 2\) | M1 | Adding circumference of two semicircles |
| \(\dfrac{28\pi}{3}\) | A1 | Must be a single term |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Area sector \(CPQ = \frac{1}{2} \times 4^2 \times \frac{\pi}{3} = \frac{8\pi}{3}\) | M1 | Uses correct formula for sector |
| Area of segment of large circle beyond \(CPQ\) \(= \frac{8\pi}{3} - \frac{1}{2} \times 4^2 \times \sin\left(\frac{\pi}{3}\right) = \frac{8\pi}{3} - 4\sqrt{3}\) | M1 | Attempts to find area of segment |
| Area of small semicircle \(= \pi \times 2\) or area of small circle \(= \pi \times 2^2\) | M1 | |
| Area of plate = Large circle \(- [2\times]\) small semicircle \(- [2\times]\) segment area | M1 | |
| \(\pi \times 4^2 - \pi \times 2^2 - 2\times\left(\frac{8\pi}{3} - 4\sqrt{3}\right) = \frac{20\pi}{3} + 8\sqrt{3}\) | A1 | AG |
| Alternative method: | ||
| Area of sector \(PCS = \frac{1}{2} \times 4^2 \times \frac{2\pi}{3} = \frac{16\pi}{3}\) | M1 | Uses correct formula for sector |
| Area of triangle \(PCQ = \frac{1}{2} \times 4^2 \times \sin\frac{\pi}{3} = 4\sqrt{3}\) | M1 | Uses correct formula for triangle |
| Area of small semicircle \(= \pi \times 2\) or area of circle \(= \pi \times 2^2\) | M1 | |
| Area of plate \(= [2\times]\) large sector \(+ [2\times]\) triangle \(- [2\times]\) small semicircle | M1 | |
| \(2\left(\frac{16\pi}{3}\right) + 2(4\sqrt{3}) - \pi \times 2^2 = \frac{20\pi}{3} + 8\sqrt{3}\) | A1 | AG |
| 5 |
## Question 8(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **Either** Let midpoint of $PQ$ be $H$: $\sin HCP = \dfrac{2}{4} \Rightarrow$ Angle $HCP = \dfrac{\pi}{6}$ | M1 | |
| **Or** $\sin PSQ = \dfrac{4}{8} \Rightarrow$ Angle $PSQ = \dfrac{\pi}{6}$ | | |
| **Or** using cosine rule: angle $PCQ = \dfrac{\pi}{3}$ | | |
| **Or** by inspection: triangle $PCQ$ or $PCT$ is equilateral so angle $PCQ = \dfrac{\pi}{3}$ | | |
| Angle $PCS = \pi - \dfrac{\pi}{6} - \dfrac{\pi}{6} = \dfrac{2}{3}\pi$ | A1 | AG |
| | **2** | |
## Question 8(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Perimeter $= 2 \times 4 \times \dfrac{2\pi}{3}$ or $8\pi - \dfrac{8\pi}{3}$ | M1 | Length of two arcs $PS$ and $QR$ |
| $+2\pi \times 2$ | M1 | Adding circumference of two semicircles |
| $\dfrac{28\pi}{3}$ | A1 | Must be a single term |
| | **3** | |
## Question 8(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Area sector $CPQ = \frac{1}{2} \times 4^2 \times \frac{\pi}{3} = \frac{8\pi}{3}$ | M1 | Uses correct formula for sector |
| Area of segment of large circle beyond $CPQ$ $= \frac{8\pi}{3} - \frac{1}{2} \times 4^2 \times \sin\left(\frac{\pi}{3}\right) = \frac{8\pi}{3} - 4\sqrt{3}$ | M1 | Attempts to find area of segment |
| Area of small semicircle $= \pi \times 2$ or area of small circle $= \pi \times 2^2$ | M1 | |
| Area of plate = Large circle $- [2\times]$ small semicircle $- [2\times]$ segment area | M1 | |
| $\pi \times 4^2 - \pi \times 2^2 - 2\times\left(\frac{8\pi}{3} - 4\sqrt{3}\right) = \frac{20\pi}{3} + 8\sqrt{3}$ | A1 | AG |
| **Alternative method:** | | |
| Area of sector $PCS = \frac{1}{2} \times 4^2 \times \frac{2\pi}{3} = \frac{16\pi}{3}$ | M1 | Uses correct formula for sector |
| Area of triangle $PCQ = \frac{1}{2} \times 4^2 \times \sin\frac{\pi}{3} = 4\sqrt{3}$ | M1 | Uses correct formula for triangle |
| Area of small semicircle $= \pi \times 2$ or area of circle $= \pi \times 2^2$ | M1 | |
| Area of plate $= [2\times]$ large sector $+ [2\times]$ triangle $- [2\times]$ small semicircle | M1 | |
| $2\left(\frac{16\pi}{3}\right) + 2(4\sqrt{3}) - \pi \times 2^2 = \frac{20\pi}{3} + 8\sqrt{3}$ | A1 | AG |
| | **5** | |
---8\\
\includegraphics[max width=\textwidth, alt={}, center]{80a20f05-61db-42d9-b4ba-53eea2290b2d-10_780_814_264_662}
The diagram shows a symmetrical metal plate. The plate is made by removing two identical pieces from a circular disc with centre $C$. The boundary of the plate consists of two $\operatorname { arcs } P S$ and $Q R$ of the original circle and two semicircles with $P Q$ and $R S$ as diameters. The radius of the circle with centre $C$ is 4 cm , and $P Q = R S = 4 \mathrm {~cm}$ also.
\begin{enumerate}[label=(\alph*)]
\item Show that angle $P C S = \frac { 2 } { 3 } \pi$ radians.
\item Find the exact perimeter of the plate.
\item Show that the area of the plate is $\left( \frac { 20 } { 3 } \pi + 8 \sqrt { 3 } \right) \mathrm { cm } ^ { 2 }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2021 Q8 [10]}}