| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2024 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Multiple circles or sectors |
| Difficulty | Standard +0.3 This is a standard multi-part sector problem requiring arc length and area formulas with radians. Part (a) involves setting up equations using the given ratio and arc length to find a radius (straightforward algebra). Parts (b) and (c) apply standard formulas for perimeter and area of sectors. While it requires careful organization across multiple steps, all techniques are routine P1 content with no novel insight needed—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/Working | Mark | Guidance |
| Attempts \(S = r\theta \Rightarrow 9 = OD \times 0.8 \Rightarrow OD = 11.25\) or \(\frac{45}{4}\) | M1, A1 | Attempts to use \(S = r\theta\); implied by \((r) = \frac{9}{0.8}\) |
| \(AO = \frac{5}{8} \times 11.25 = 7.03\) m | A1* | Shows \(AO\) is \(7.03(m)\) via \(\frac{5}{8}\times11.25\), \(11.25 - \frac{3}{8}\times11.25\). Units not required. Achieves \(OD=11.25\) or \(OC=11.25\); condone \(r=11.25\). May be implied by \(AO=\frac{5}{8}\times\left(\frac{9}{0.8}\right)\) |
| (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts \(A = 7.03\times(2\pi - 0.8) = (38.55)\) | M1 | Attempts arc length \(AEB\) via \(7.03\times(2\pi-0.8)\); allow angle implied by sight of awrt 5.48 |
| Attempts \(9 + 2\times(11.25 - 7.03) + 7.03\times\theta\) | M1 | Attempts to add correct parts together e.g. \(9 + 2\times(``11.25''-7.03)+7.03\times\theta\); allow incorrect angle for candidates who don't equate \(2\pi\) to \(360°\) |
| Perimeter \(=\) awrt \(56.0\) m | A1 | ISW after correct answer; condone 56 m following correct calculation; units not necessary |
| (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts \(\frac{1}{2}\times``11.25''^2\times0.8=(50.625)\) OR \(\frac{1}{2}\times7.03^2\times(2\pi-0.8)=(135.4)\) | M1 | Attempts \(\frac{1}{2}r^2\theta\) with \(r=11.25\), \(\theta=0.8\) or \(r=7.03\), \(\theta=2\pi-0.8\) or awrt 5.48; alt for second sector: \(\pi\times7.03^2 - \frac{1}{2}\times7.03^2\times0.8\) |
| Full method: \(\frac{1}{2}\times``11.25''^2\times0.8 + \frac{1}{2}\times7.03^2\times(2\pi-0.8)\) | dM1 | Adds two sectors |
| \(=\) awrt \(186\ \text{m}^2\) | A1 | Units not necessary |
| (3 marks) |
# Question 8:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $S = r\theta \Rightarrow 9 = OD \times 0.8 \Rightarrow OD = 11.25$ or $\frac{45}{4}$ | M1, A1 | Attempts to use $S = r\theta$; implied by $(r) = \frac{9}{0.8}$ |
| $AO = \frac{5}{8} \times 11.25 = 7.03$ m | A1* | Shows $AO$ is $7.03(m)$ via $\frac{5}{8}\times11.25$, $11.25 - \frac{3}{8}\times11.25$. Units not required. Achieves $OD=11.25$ or $OC=11.25$; condone $r=11.25$. May be implied by $AO=\frac{5}{8}\times\left(\frac{9}{0.8}\right)$ |
| **(3 marks)** | | |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $A = 7.03\times(2\pi - 0.8) = (38.55)$ | M1 | Attempts arc length $AEB$ via $7.03\times(2\pi-0.8)$; allow angle implied by sight of awrt 5.48 |
| Attempts $9 + 2\times(11.25 - 7.03) + 7.03\times\theta$ | M1 | Attempts to add correct parts together e.g. $9 + 2\times(``11.25''-7.03)+7.03\times\theta$; allow incorrect angle for candidates who don't equate $2\pi$ to $360°$ |
| Perimeter $=$ awrt $56.0$ m | A1 | ISW after correct answer; condone 56 m following correct calculation; units not necessary |
| **(3 marks)** | | |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $\frac{1}{2}\times``11.25''^2\times0.8=(50.625)$ OR $\frac{1}{2}\times7.03^2\times(2\pi-0.8)=(135.4)$ | M1 | Attempts $\frac{1}{2}r^2\theta$ with $r=11.25$, $\theta=0.8$ or $r=7.03$, $\theta=2\pi-0.8$ or awrt 5.48; alt for second sector: $\pi\times7.03^2 - \frac{1}{2}\times7.03^2\times0.8$ |
| Full method: $\frac{1}{2}\times``11.25''^2\times0.8 + \frac{1}{2}\times7.03^2\times(2\pi-0.8)$ | dM1 | Adds two sectors |
| $=$ awrt $186\ \text{m}^2$ | A1 | Units not necessary |
| **(3 marks)** | | |
---8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2043b938-ed3f-4b69-9ea9-b4ab62e2a8ce-18_680_933_294_589}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a sketch of the plan view of a platform.\\
The plan view of the platform consists of a sector $D O C$ of a circle centre $O$ joined to a sector $A O B E A$ of a different circle, also with centre $O$.
Given that
\begin{itemize}
\item angle $A O B = 0.8$ radians
\item arc length $C D = 9 \mathrm {~m}$
\item $D A : A O = 3 : 5$
\begin{enumerate}[label=(\alph*)]
\item show that $A O = 7.03 \mathrm {~m}$ to 3 significant figures.
\item Find the perimeter of the platform, in m , to 3 significant figures.
\item Find the total area of the platform, giving your answer in $\mathrm { m } ^ { 2 }$ to the nearest whole number.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2024 Q8 [9]}}