| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | October |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Sector with attached triangle |
| Difficulty | Moderate -0.3 This is a multi-part question testing standard radian formulas (sector area, arc length) and basic trigonometry (triangle area, cosine rule, sine rule). While it has multiple parts (a-d), each step uses routine formulas with no novel problem-solving required. The 'show that' in part (c) and finding the quadrilateral area in (d) add slight complexity, but overall this is slightly easier than average due to its formulaic nature and clear structure. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.02i Represent inequalities: graphically on coordinate plane1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2\pi - \frac{2\pi}{3} = \frac{4\pi}{3}\) | B1 | Finds correct angle for sector AOBX; may find minor sector first then subtract; sight of \(\theta = \frac{4\pi}{3}\) on diagram or in working scores this mark |
| Area of sector \(= \frac{1}{2} \times 3^2 \times \frac{4}{3}\pi = 6\pi\) (m²) | M1A1 | M1: states or uses \(\frac{1}{2}r^2\theta\) with \(r=3\) and \(\theta = \frac{2\pi}{3}\) or \(\theta = \frac{4\pi}{3}\); A1: \(6\pi\) (m²) cao, must be exact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Length of arc \(= 3 \times \frac{4}{3}\pi \Rightarrow\) Perimeter \(= 4\pi + 6\) (m) | M1A1 | M1: states or uses \(r\theta\) with \(r=3\) and \(\theta=\frac{2\pi}{3}\) or \(\frac{4\pi}{3}\); addition of two radii not required for M1; A1: \(4\pi+6\) (m) cao, must be exact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2} \times 3^2 \times \sin\!\left(\frac{2}{3}\pi\right) = \frac{9\sqrt{3}}{4}\) (m²) | M1A1 | M1: states or uses \(\frac{1}{2}ab\sin C\) with \(a=b=3\), \(\theta=\frac{2}{3}\pi\); may split into two right-angled triangles; score for overall method or awrt 3.90; A1: \(\frac{9\sqrt{3}}{4}\) (m²) or exact equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(AB^2 = 3^2 + 3^2 - 2\times3\times3\times\cos\!\left(\frac{2\pi}{3}\right) \Rightarrow AB^2 = 27 \Rightarrow AB = 3\sqrt{3}\) (m) | M1A1* | M1: correct method e.g. cosine rule with \(a=b=3\), \(\theta=\frac{2}{3}\pi\); or splitting isosceles triangle; or sine rule \(\frac{AB}{\sin(2\pi/3)}=\frac{3}{\sin(\pi/6)}\); A1*: \(3\sqrt{3}\) (m) with no errors, minimum simplified expression seen e.g. \(AB^2=27\) or \(AB=\sqrt{27}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{\sin BAC}{8} = \frac{\sin\!\left(\frac{\pi}{6}\right)}{3\sqrt{3}} \Rightarrow \sin BAC = \ldots\) or \(BAC = \ldots\) | M1 | Attempts sine rule to find \(\sin BAC\) or angle \(BAC\); award for appropriate lengths and angles in correct positions; alternatively attempts cosine rule quadratic in \(AC\) |
| \(\sin BAC = \text{awrt } \frac{4\sqrt{3}}{9}\) or \(BAC = \text{awrt } 0.88\) (0.8785…) | A1 | Correct value for \(\sin BAC\), angle \(BAC\) (awrt 0.88 or 50.3°) or \(AC\) |
| Area \(ABC = \frac{1}{2}\times3\sqrt{3}\times8\times\sin\!\left(\pi - \frac{\pi}{6} - \text{"0.88"}\right)\) (= 20.4896…) | M1 | Correct method to find area of triangle \(ABC\) using angle \(BAC\) from \(\pi - \frac{\pi}{6} - \text{"BAC"}\) or their length \(AC\) |
| Total area \(=\) "18.8" \(+\) "3.90" \(+\) "20.5" \(= \text{awrt } 43\) (m²) | dM1 | Adds (a)(i) + (b) + triangle \(ABC\); dependent on all previous method marks in (d) |
| A1 | awrt 43 (m²) from correct method; exact answer \(6\pi + 2\sqrt{11} + \frac{41}{4}\sqrt{3}\) also accepted |
## Question 8:
### Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2\pi - \frac{2\pi}{3} = \frac{4\pi}{3}$ | B1 | Finds correct angle for sector AOBX; may find minor sector first then subtract; sight of $\theta = \frac{4\pi}{3}$ on diagram or in working scores this mark |
| Area of sector $= \frac{1}{2} \times 3^2 \times \frac{4}{3}\pi = 6\pi$ (m²) | M1A1 | M1: states or uses $\frac{1}{2}r^2\theta$ with $r=3$ and $\theta = \frac{2\pi}{3}$ or $\theta = \frac{4\pi}{3}$; A1: $6\pi$ (m²) cao, must be exact |
### Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Length of arc $= 3 \times \frac{4}{3}\pi \Rightarrow$ Perimeter $= 4\pi + 6$ (m) | M1A1 | M1: states or uses $r\theta$ with $r=3$ and $\theta=\frac{2\pi}{3}$ or $\frac{4\pi}{3}$; addition of two radii not required for M1; A1: $4\pi+6$ (m) cao, must be exact |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2} \times 3^2 \times \sin\!\left(\frac{2}{3}\pi\right) = \frac{9\sqrt{3}}{4}$ (m²) | M1A1 | M1: states or uses $\frac{1}{2}ab\sin C$ with $a=b=3$, $\theta=\frac{2}{3}\pi$; may split into two right-angled triangles; score for overall method or awrt 3.90; A1: $\frac{9\sqrt{3}}{4}$ (m²) or exact equivalent |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $AB^2 = 3^2 + 3^2 - 2\times3\times3\times\cos\!\left(\frac{2\pi}{3}\right) \Rightarrow AB^2 = 27 \Rightarrow AB = 3\sqrt{3}$ (m) | M1A1* | M1: correct method e.g. cosine rule with $a=b=3$, $\theta=\frac{2}{3}\pi$; or splitting isosceles triangle; or sine rule $\frac{AB}{\sin(2\pi/3)}=\frac{3}{\sin(\pi/6)}$; A1*: $3\sqrt{3}$ (m) with no errors, minimum simplified expression seen e.g. $AB^2=27$ or $AB=\sqrt{27}$ |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{\sin BAC}{8} = \frac{\sin\!\left(\frac{\pi}{6}\right)}{3\sqrt{3}} \Rightarrow \sin BAC = \ldots$ or $BAC = \ldots$ | M1 | Attempts sine rule to find $\sin BAC$ or angle $BAC$; award for appropriate lengths and angles in correct positions; alternatively attempts cosine rule quadratic in $AC$ |
| $\sin BAC = \text{awrt } \frac{4\sqrt{3}}{9}$ or $BAC = \text{awrt } 0.88$ (0.8785…) | A1 | Correct value for $\sin BAC$, angle $BAC$ (awrt 0.88 or 50.3°) or $AC$ |
| Area $ABC = \frac{1}{2}\times3\sqrt{3}\times8\times\sin\!\left(\pi - \frac{\pi}{6} - \text{"0.88"}\right)$ (= 20.4896…) | M1 | Correct method to find area of triangle $ABC$ using angle $BAC$ from $\pi - \frac{\pi}{6} - \text{"BAC"}$ or their length $AC$ |
| Total area $=$ "18.8" $+$ "3.90" $+$ "20.5" $= \text{awrt } 43$ (m²) | dM1 | Adds (a)(i) + (b) + triangle $ABC$; dependent on all previous method marks in (d) |
| | A1 | awrt 43 (m²) from correct method; exact answer $6\pi + 2\sqrt{11} + \frac{41}{4}\sqrt{3}$ also accepted |
---8.
\section*{Diagram NOT to scale}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{db979349-3415-420f-a39f-8cc8c24a69d0-20_461_1036_296_534}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows the plan view of a design for a pond.\\
The design consists of a sector $A O B X$ of a circle centre $O$ joined to a quadrilateral $A O B C$.
\begin{itemize}
\item $B C = 8 \mathrm {~m}$
\item $O A = O B = 3 \mathrm {~m}$
\item angle $A O B$ is $\frac { 2 \pi } { 3 }$ radians
\item angle $B C A$ is $\frac { \pi } { 6 }$ radians
\begin{enumerate}[label=(\alph*)]
\item Calculate (i) the exact area of the sector $A O B X$,\\
(ii) the exact perimeter of the sector $A O B X$.
\item Calculate the exact area of the triangle $A O B$.
\item Show that the length $A B$ is $3 \sqrt { 3 } \mathrm {~m}$.
\item Find the total surface area of the pond. Give your answer in $\mathrm { m } ^ { 2 }$ correct to 2 significant figures.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2022 Q8 [14]}}