| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2020 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Sector and arc length |
| Difficulty | Moderate -0.5 This is a straightforward application of circle geometry, arc length, and sector area formulas. Part (a) uses inverse trigonometry (arcsin or arccos) with given dimensions, while parts (b) and (c) apply standard sector formulas. The multi-part structure and geometric setup add slight complexity, but all techniques are routine for P1 level with no novel problem-solving required. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (a) \(\cos AOD = \frac{4}{12} \Rightarrow\) angle \(AOD = 1.231\) | M1 A1* | M1: Attempts \(\cos AOD = \frac{4}{12}\) via correct method. A1*: Achieves angle \(AOD = 1.231\) following valid method with at least one step shown. Withhold if awrt 1.231 not achieved by correct method, or if working in degrees without achieving awrt \(70.5°\) before converting. |
| (b) Attempts \(\frac{1}{2}r^2\theta\) with \(r=12\), \(\theta = \pi \pm 1.231\) or \(1.231\) | M1 | \(\angle AOC =\) awrt \(1.911\) \((\pi - 1.231)\) or \(\angle AOC =\) awrt \(4.373\) \((\pi + 1.231)\) |
| Attempts area \(AOD = \frac{1}{2} \times\)"\(4\)"\(\times\sqrt{12^2 -\text{"\)4\("}^2}\) | M1 | Angle may be in degrees \((70.5°...)\). Allow errors in finding "4" but \(r\) must be 12. |
| Attempts sector \(-\) triangle \(= \frac{1}{2}\times12^2\times(\pi+1.231) - \frac{1}{2}\times\)"\(4\)"\(\times\sqrt{12^2-\text{"\)4\("}^2}\) giving \((314.8...) - (22.627...)\) | ddM1 | Full method, dependent on both previous M marks. Angle may be awrt 4.4. |
| \(=\) awrt \(292.2\ (\text{m}^2)\) | A1 | |
| (c) Attempts \(s = r\theta\) with \(r=12\), \(\theta = \pi \pm 1.231\) or \(1.231\) | M1 | Embedded values sufficient. Condone rounding of 1.231 or use of awrt 4.4 or awrt 1.9 radians. |
| Attempts \(P = 16 + \sqrt{12^2-4^2} + 12(\pi+1.231)\) | dM1 | Full method for perimeter. Angle may be awrt 4.4 or awrt 1.9. |
| \(=\) awrt \(79.8\ (\text{m})\) | A1 |
## Question 4:
| Answer/Working | Marks | Guidance |
|---|---|---|
| **(a)** $\cos AOD = \frac{4}{12} \Rightarrow$ angle $AOD = 1.231$ | M1 A1* | M1: Attempts $\cos AOD = \frac{4}{12}$ via correct method. A1*: Achieves angle $AOD = 1.231$ following valid method with at least one step shown. Withhold if awrt 1.231 not achieved by correct method, or if working in degrees without achieving awrt $70.5°$ before converting. |
| **(b)** Attempts $\frac{1}{2}r^2\theta$ with $r=12$, $\theta = \pi \pm 1.231$ or $1.231$ | M1 | $\angle AOC =$ awrt $1.911$ $(\pi - 1.231)$ or $\angle AOC =$ awrt $4.373$ $(\pi + 1.231)$ |
| Attempts area $AOD = \frac{1}{2} \times$"$4$"$\times\sqrt{12^2 -\text{"$4$"}^2}$ | M1 | Angle may be in degrees $(70.5°...)$. Allow errors in finding "4" but $r$ must be 12. |
| Attempts sector $-$ triangle $= \frac{1}{2}\times12^2\times(\pi+1.231) - \frac{1}{2}\times$"$4$"$\times\sqrt{12^2-\text{"$4$"}^2}$ giving $(314.8...) - (22.627...)$ | ddM1 | Full method, dependent on both previous M marks. Angle may be awrt 4.4. |
| $=$ awrt $292.2\ (\text{m}^2)$ | A1 | |
| **(c)** Attempts $s = r\theta$ with $r=12$, $\theta = \pi \pm 1.231$ or $1.231$ | M1 | Embedded values sufficient. Condone rounding of 1.231 or use of awrt 4.4 or awrt 1.9 radians. |
| Attempts $P = 16 + \sqrt{12^2-4^2} + 12(\pi+1.231)$ | dM1 | Full method for perimeter. Angle may be awrt 4.4 or awrt 1.9. |
| $=$ awrt $79.8\ (\text{m})$ | A1 | |4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{28839dd5-b9c1-4cbd-981e-8f79c43ba086-08_622_894_258_683}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows the plan view of a house $A B C D$ and a lawn $A P C D A$.\\
$A B C D$ is a rectangle with $A B = 16 \mathrm {~m}$.\\
$A P C O A$ is a sector of a circle centre $O$ with radius 12 m .
The point $O$ lies on the line $D C$, as shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Show that the size of angle $A O D$ is 1.231 radians to 3 decimal places.
The lawn $A P C D A$ is shown shaded in Figure 2.
\item Find the area of the lawn, in $\mathrm { m } ^ { 2 }$, to one decimal place.
\item Find the perimeter of the lawn, in metres, to one decimal place.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2020 Q4 [9]}}