| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Sector with attached triangle |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on sector area and arc length formulas with basic trigonometry. Part (a) uses the sector area formula directly, (b) requires simple right-angled triangle trigonometry, and (c)-(d) involve adding areas and perimeters. All steps are routine applications of standard formulas with no problem-solving insight 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 |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(A = \frac{1}{2}r^2\theta \Rightarrow 40 = \frac{1}{2}r^2 \times 2.4 \Rightarrow r = \ldots\) | M1 | Attempts to use \(A = \frac{1}{2}r^2\theta\) with \(A=40\) and \(\theta = 2.4\) |
| \(r = \sqrt{\frac{80}{2.4}} \Rightarrow r = 5.77\) (m) | A1 | Achieves awrt 5.77 (m) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Angle \(AOB\) = awrt \(0.37\) | B1 | Allow this to score anywhere in their answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{1}{2} \times\) "5.77"\( \times 6.25 \times \sin\)"\(0.37\)" \((= 6.5377\ldots)\) | M1 | Attempts area of triangle \(AOB\) (or \(DOC\)). May be seen "doubled". NB "doubled" is 13.075… |
| Full method for area of stage \(= 40 + 2 \times \frac{1}{2} \times 5.77 \times 6.25 \times \sin 0.37\) | dM1 | Attempts area of stage \(40 + 2 \times \frac{1}{2} \times\)"\(a\)"\(\times 6.25 \times \sin\)"\(b\)" or e.g. \(\frac{1}{2} \times\)"\(5.77\)"\(^2 \times 2.4 + 2 \times \frac{1}{2} \times\)"\(a\)"\(\times 6.25 \times \sin\)"\(b\)" |
| \(= 53.1 \text{ m}^2\) | A1 | Awrt 53.1 (m²). Condone awrt 53.0 but not just 53 unless awrt 53.1 or 53.0 seen earlier |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(r\theta =\) "5.77"\(\times 2.4 = (13.848)\) | M1 | Attempts \(r\theta =\) "\(a\)"\(\times 2.4\) |
| \(x^2 = 6.25^2 +\) "5.77"\(^2 - 2 \times 6.25 \times\)"5.77"\(\cos\)"0.37"\( \quad (x = 2.26)\) | M1 | Attempts cosine rule to find length \(AB\) or \(AB^2\) (or \(CD/CD^2\)). Allow anywhere in solution |
| Full method for perimeter of stage \(= 12.5 + 2 \times\)"2.26"\(+\)"5.77"\(\times 2.4\) | ddM1 | Full method to find perimeter \(12.5 + 2 \times AB +\)"\(a\)"\(\times 2.4\). Must use \(AB\) not \(AB^2\). Depends on both previous M marks |
| \(= 30.9\) m | A1 | Awrt 30.9 (m). Beware: \(AB = 6.25\sin(0.37) = 2.26\ldots\) erroneously leads to correct answer — scores M1M0ddM1A0 if arc length correct |
## Question 6(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $A = \frac{1}{2}r^2\theta \Rightarrow 40 = \frac{1}{2}r^2 \times 2.4 \Rightarrow r = \ldots$ | M1 | Attempts to use $A = \frac{1}{2}r^2\theta$ with $A=40$ and $\theta = 2.4$ |
| $r = \sqrt{\frac{80}{2.4}} \Rightarrow r = 5.77$ (m) | A1 | Achieves awrt 5.77 (m) |
---
## Question 6(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Angle $AOB$ = awrt $0.37$ | B1 | Allow this to score anywhere in their answer |
---
## Question 6(c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{1}{2} \times$ "5.77"$ \times 6.25 \times \sin$"$0.37$" $(= 6.5377\ldots)$ | M1 | Attempts area of triangle $AOB$ (or $DOC$). May be seen "doubled". NB "doubled" is 13.075… |
| Full method for area of stage $= 40 + 2 \times \frac{1}{2} \times 5.77 \times 6.25 \times \sin 0.37$ | dM1 | Attempts area of stage $40 + 2 \times \frac{1}{2} \times$"$a$"$\times 6.25 \times \sin$"$b$" or e.g. $\frac{1}{2} \times$"$5.77$"$^2 \times 2.4 + 2 \times \frac{1}{2} \times$"$a$"$\times 6.25 \times \sin$"$b$" |
| $= 53.1 \text{ m}^2$ | A1 | Awrt 53.1 (m²). Condone awrt 53.0 but not just 53 unless awrt 53.1 or 53.0 seen earlier |
---
## Question 6(d):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $r\theta =$ "5.77"$\times 2.4 = (13.848)$ | M1 | Attempts $r\theta =$ "$a$"$\times 2.4$ |
| $x^2 = 6.25^2 +$ "5.77"$^2 - 2 \times 6.25 \times$"5.77"$\cos$"0.37"$ \quad (x = 2.26)$ | M1 | Attempts cosine rule to find length $AB$ or $AB^2$ (or $CD/CD^2$). Allow anywhere in solution |
| Full method for perimeter of stage $= 12.5 + 2 \times$"2.26"$+$"5.77"$\times 2.4$ | ddM1 | Full method to find perimeter $12.5 + 2 \times AB +$"$a$"$\times 2.4$. Must use $AB$ not $AB^2$. Depends on both previous M marks |
| $= 30.9$ m | A1 | Awrt 30.9 (m). Beware: $AB = 6.25\sin(0.37) = 2.26\ldots$ erroneously leads to correct answer — scores M1M0ddM1A0 if arc length correct |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{bb21001f-fe68-4776-992d-ede1aae233d7-12_438_816_246_621}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Diagram NOT accurately drawn
Figure 1 shows the plan view for the design of a stage.\\
The design consists of a sector $O B C$ of a circle, with centre $O$, joined to two congruent triangles $O A B$ and $O D C$.
Given that
\begin{itemize}
\item angle $B O C = 2.4$ radians
\item area of sector $B O C = 40 \mathrm {~m} ^ { 2 }$
\item $A O D$ is a straight line of length 12.5 m
\begin{enumerate}[label=(\alph*)]
\item find the radius of the sector, giving your answer, in m , to 2 decimal places,
\item find the size of angle $A O B$, in radians, to 2 decimal places.
\end{itemize}
Hence find
\item the total area of the stage, giving your answer, in $\mathrm { m } ^ { 2 }$, to one decimal place,
\item the total perimeter of the stage, giving your answer, in m , to one decimal place.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2023 Q6 [10]}}