| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| 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 straightforward application of arc length and sector area formulas with clear given information. Students must use s=rθ to find the angle, then apply it to find another arc length, perimeter, and area. The multi-part structure and need to work with multiple sectors adds slight complexity, but all steps follow standard procedures without requiring problem-solving insight or novel approaches. |
| 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 |
| \(15 = 9 \times \theta \Rightarrow \theta = \frac{5}{3} = (1.67)\) | M1 | Correct use of arc length formula to find angle subtended by arc \(AB\); attempts \(15 = 9 \times \theta \Rightarrow \theta = ...\) |
| \(\phi = \frac{2\pi}{3} - \frac{5}{3} = (0.4277...)\) | dM1 | Correct method to find angle subtended by arc \(CD\) using their angle for arc \(AB\); note \(\phi = \frac{1}{3}\left(2\pi - 3 \times \frac{5}{3}\right)\) also correct; dependent on previous M |
| Arc \(CD = 84 \times \left(\frac{2\pi}{3} - \frac{5}{3}\right) = 35.929... = 35.9\) cm (1 d.p.) | A1* | CSO; arrives at 35.9 with correct value to at least 2 d.p. seen first; alternatively sight of \(84 \times \left(\frac{2\pi}{3} - \frac{5}{3}\right)\) or \(84 \times\) awrt 0.4277 followed by 35.9 cm |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Perimeter \(= 3 \times (15 + 35.9...) + 6 \times (84 - 9)\) | M1 | Correct method for perimeter; must include all six arcs and radial edges; look for \(3\times15 + 3\times35.9 + 6\times...\); implied by awrt 603 |
| \(=\) awrt 603 cm \((602.787...)\) | A1 | Units need not be given |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Area of a "blade" \(= \frac{1}{2} \times 84^2 \times \left(\frac{2\pi}{3} - \frac{5}{3}\right) =\) awrt (1510) | M1 | A correct attempt at any relevant area |
| Area of sector of inner circle between blades \(= \frac{1}{2} \times 9^2 \times \frac{5}{3} = (67.5)\) | dM1 A1 | A correct attempt at a corresponding area combinable with first area; both areas correct; angles need not be calculated but must be correct to 3sf; dependent on previous M |
| Total area \(= 3\left(\frac{1}{2}\times84^2\times\left(\frac{2\pi}{3}-\frac{5}{3}\right)+\frac{1}{2}\times9^2\times\frac{5}{3}\right) = ...(4729.577764\text{ cm}^2)\) | ddM1 | Correct combination of areas to find area of fan |
| Area is awrt \(0.473\) m\(^2\) or awrt \(4730\) cm\(^2\) | A1 | Must include units; ISW after correct answer |
## Question 8:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $15 = 9 \times \theta \Rightarrow \theta = \frac{5}{3} = (1.67)$ | M1 | Correct use of arc length formula to find angle subtended by arc $AB$; attempts $15 = 9 \times \theta \Rightarrow \theta = ...$ |
| $\phi = \frac{2\pi}{3} - \frac{5}{3} = (0.4277...)$ | dM1 | Correct method to find angle subtended by arc $CD$ using their angle for arc $AB$; note $\phi = \frac{1}{3}\left(2\pi - 3 \times \frac{5}{3}\right)$ also correct; dependent on previous M |
| Arc $CD = 84 \times \left(\frac{2\pi}{3} - \frac{5}{3}\right) = 35.929... = 35.9$ cm (1 d.p.) | A1* | CSO; arrives at 35.9 with correct value to at least 2 d.p. seen first; alternatively sight of $84 \times \left(\frac{2\pi}{3} - \frac{5}{3}\right)$ or $84 \times$ awrt 0.4277 followed by 35.9 cm |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Perimeter $= 3 \times (15 + 35.9...) + 6 \times (84 - 9)$ | M1 | Correct method for perimeter; must include all six arcs and radial edges; look for $3\times15 + 3\times35.9 + 6\times...$; implied by awrt 603 |
| $=$ awrt 603 cm $(602.787...)$ | A1 | Units need not be given |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Area of a "blade" $= \frac{1}{2} \times 84^2 \times \left(\frac{2\pi}{3} - \frac{5}{3}\right) =$ awrt (1510) | M1 | A correct attempt at any relevant area |
| Area of sector of inner circle between blades $= \frac{1}{2} \times 9^2 \times \frac{5}{3} = (67.5)$ | dM1 A1 | A correct attempt at a corresponding area combinable with first area; both areas correct; angles need not be calculated but must be correct to 3sf; dependent on previous M |
| Total area $= 3\left(\frac{1}{2}\times84^2\times\left(\frac{2\pi}{3}-\frac{5}{3}\right)+\frac{1}{2}\times9^2\times\frac{5}{3}\right) = ...(4729.577764\text{ cm}^2)$ | ddM1 | Correct combination of areas to find area of fan |
| Area is awrt $0.473$ m$^2$ or awrt $4730$ cm$^2$ | A1 | Must include units; ISW after correct answer |
---8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3cf69966-e825-4ff0-a6e8-c5dfdc92c53f-22_922_876_246_539}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a sketch of the outline of the face of a ceiling fan viewed from below.\\
The fan consists of three identical sections congruent to $O A B C D O$, shown in Figure 3, where
\begin{itemize}
\item $O A B O$ is a sector of a circle with centre $O$ and radius 9 cm
\item $O B C D O$ is a sector of a circle with centre $O$ and radius 84 cm
\item angle $A O D = \frac { 2 \pi } { 3 }$ radians
\end{itemize}
Given that the length of the arc $A B$ is 15 cm ,
\begin{enumerate}[label=(\alph*)]
\item show that the length of the arc $C D$ is 35.9 cm to one decimal place.
The face of the fan is modelled to be a flat surface.\\
Find, according to the model,
\item the perimeter of the face of the fan, giving your answer to the nearest cm,
\item the surface area of the face of the fan.
Give your answer to 3 significant figures and make your units clear.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2022 Q8 [10]}}