Edexcel Paper 1 2018 June — Question 3 4 marks

Exam BoardEdexcel
ModulePaper 1 (Paper 1)
Year2018
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRadians, Arc Length and Sector Area
TypeSimultaneous equations with arc/area
DifficultyStandard +0.3 This question requires setting up two simultaneous equations from the given conditions (area = 11 and perimeter = 4 × arc length) using standard formulas (area = ½r²θ, arc = rθ, perimeter = 2r + rθ). The algebra is straightforward once the equations are established, making this slightly easier than average for A-level.
Spec1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta

3. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{b5f50f17-9f1b-4b4c-baf3-e50de5f2ea9c-06_332_348_246_861} \captionsetup{labelformat=empty} \caption{Figure 1}
\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 \(\theta\) radians.
The area of the sector \(A O B\) is \(11 \mathrm {~cm} ^ { 2 }\) Given that the perimeter of the sector is 4 times the length of the arc \(A B\), find the exact value of \(r\).

Question 3:
AnswerMarks Guidance
Working/AnswerMark Guidance
States or uses \(\frac{1}{2}r^2\theta = 11\)B1 1.1b
States or uses \(2r + r\theta = 4r\theta\)B1 1.1b
Attempts to solve, full method \(r = \ldots\)M1 3.1a
\(r = \sqrt{33}\)A1 1.1b
Notes: B1: \(\frac{1}{2}r^2\theta = 11\) may be implied by embedded value for \(\theta\). B1: \(2r + r\theta = 4r\theta\) or equivalent. M1: Full method combining equations to eliminate \(\theta\) or find \(\theta\). Cannot be scored from impossible values of \(\theta\); only score if \(0 < \theta < 2\pi\). FYI \(\theta = \frac{2}{3}\) radians. A1: \(r = \sqrt{33}\) only, isw after correct answer.
# Question 3:

| Working/Answer | Mark | Guidance |
|---|---|---|
| States or uses $\frac{1}{2}r^2\theta = 11$ | B1 | 1.1b |
| States or uses $2r + r\theta = 4r\theta$ | B1 | 1.1b |
| Attempts to solve, full method $r = \ldots$ | M1 | 3.1a |
| $r = \sqrt{33}$ | A1 | 1.1b |

**Notes:** B1: $\frac{1}{2}r^2\theta = 11$ may be implied by embedded value for $\theta$. B1: $2r + r\theta = 4r\theta$ or equivalent. M1: Full method combining equations to eliminate $\theta$ or find $\theta$. Cannot be scored from impossible values of $\theta$; only score if $0 < \theta < 2\pi$. FYI $\theta = \frac{2}{3}$ radians. A1: $r = \sqrt{33}$ only, isw after correct answer.

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3.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{b5f50f17-9f1b-4b4c-baf3-e50de5f2ea9c-06_332_348_246_861}
\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 $\theta$ radians.\\
The area of the sector $A O B$ is $11 \mathrm {~cm} ^ { 2 }$

Given that the perimeter of the sector is 4 times the length of the arc $A B$, find the exact value of $r$.

\hfill \mbox{\textit{Edexcel Paper 1 2018 Q3 [4]}}