| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | October |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Sector area calculation |
| Difficulty | Moderate -0.8 This is a straightforward application of the sector area formula (A = ½r²θ) requiring simple algebraic manipulation to find r, followed by using the arc length formula (s = rθ). Both parts involve direct substitution into standard formulas with minimal problem-solving, making it easier than average but not trivial since it requires exact form answers. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Sets \(\frac{1}{2}r^2 \times 1.25 = 15 \Rightarrow r^2 = 24\) | M1 | Uses \(A = \frac{1}{2}r^2\theta\) in an attempt to find \(r\) |
| \(r = \sqrt{24}\) or \(2\sqrt{6}\) (only) | A1 | (oe) isw after correct answer seen. Withhold A1 if \(r = \pm2\sqrt{6}\) given |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Attempts \(s = r\theta = 2\sqrt{6} \times 1.25\) | M1 | Uses \(s = r\theta\) with their \(r\) and \(\theta = 1.25\) |
| Attempts \(P = 2r + r\theta = 2\times2\sqrt{6} + 2\sqrt{6}\times1.25\) | dM1 | For applying \(P = 2r + r\theta\) with their \(r\) and \(\theta = 1.25\) |
| \(= \frac{13\sqrt{6}}{2}\) oe | A1 | isw after correct answer seen. Accept \(6.5\sqrt{13}\) (oe simplest forms) |
# Question 1:
## Part (a)
| Working | Mark | Guidance |
|---------|------|----------|
| Sets $\frac{1}{2}r^2 \times 1.25 = 15 \Rightarrow r^2 = 24$ | M1 | Uses $A = \frac{1}{2}r^2\theta$ in an attempt to find $r$ |
| $r = \sqrt{24}$ or $2\sqrt{6}$ (only) | A1 | (oe) isw after correct answer seen. Withhold A1 if $r = \pm2\sqrt{6}$ given |
## Part (b)
| Working | Mark | Guidance |
|---------|------|----------|
| Attempts $s = r\theta = 2\sqrt{6} \times 1.25$ | M1 | Uses $s = r\theta$ with their $r$ and $\theta = 1.25$ |
| Attempts $P = 2r + r\theta = 2\times2\sqrt{6} + 2\sqrt{6}\times1.25$ | dM1 | For applying $P = 2r + r\theta$ with their $r$ and $\theta = 1.25$ |
| $= \frac{13\sqrt{6}}{2}$ oe | A1 | isw after correct answer seen. Accept $6.5\sqrt{13}$ (oe simplest forms) |
---1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{50ec901b-b6b6-4b72-85bd-a084f313c99b-02_488_376_287_790}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sector $A O B$ of a circle with centre $O$ and radius $r \mathrm {~cm}$.
The angle $A O B$ is 1.25 radians.
Given that the area of the sector $A O B$ is $15 \mathrm {~cm} ^ { 2 }$
\begin{enumerate}[label=(\alph*)]
\item find the exact value of $r$,
\item find the exact length of the perimeter of the sector. Write your answer in simplest form.\\
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2019 Q1 [5]}}