AQA Paper 1 Specimen — Question 5 8 marks

Exam BoardAQA
ModulePaper 1 (Paper 1)
SessionSpecimen
Marks8
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Mark schemeDownload PDF ↗
TopicRadians, Arc Length and Sector Area
TypeSimultaneous equations with arc/area
DifficultyModerate -0.3 This is a straightforward sector problem requiring standard formulas (area = ½r²θ, perimeter = 2r + rθ) and solving a quadratic equation. Part (a) is algebraic manipulation following clear steps, and part (b) requires checking which root gives a valid angle. While it tests multiple concepts, the approach is routine with no novel insight required, making it slightly easier than average.
Spec1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta
The diagram shows a sector \(AOB\) of a circle with centre \(O\) and radius \(r\) cm. \includegraphics{figure_5} The angle \(AOB\) is \(\theta\) radians The sector has area 9 cm\(^2\) and perimeter 15 cm.
  1. Show that \(r\) satisfies the equation \(2r^2 - 15r + 18 = 0\) [4 marks]
  2. Find the value of \(\theta\). Explain why it is the only possible value. [4 marks]
Question 5:
AnswerMarks
5(a)1
Recalls A r2 or l r
2
AnswerMarks Guidance
PI by use in equationAO1.2 B1
1 18
r29, 
2 r2
Perimeter of sector gives
2rr15
18
2r 15
r
2r2 1815r
2r215r180 (AG)
Constructs two equations at
AnswerMarks Guidance
least one correctAO1.1a M1
Eliminates 
AnswerMarks Guidance
FT incorrect equationsAO1.1a M1
Constructs a rigorous
mathematical argument to
show required result, clearly
constructing two correct
simultaneous equations and
AnswerMarks Guidance
eliminating AGAO2.1 R1
(b)Solves a quadratic equation
and finds two values of AO3.1a M1
r  ,r 6
2
1
r 6
2
3
r  8
2
828
so only one possible value of 
AnswerMarks Guidance
Finds two correct values of rAO1.1b B1
Finds both values of AO1.1b A1
Gives a valid reason for
AnswerMarks Guidance
rejecting one of ‘their’ valuesAO2.4 R1
Total8
QMarking Instructions AO 
Question 5:
--- 5(a) ---
5(a) | 1
Recalls A r2 or l r
2
PI by use in equation | AO1.2 | B1 | Area of sector gives
1 18
r29, 
2 r2
Perimeter of sector gives
2rr15
18
2r 15
r
2r2 1815r
2r215r180 (AG)
Constructs two equations at
least one correct | AO1.1a | M1
Eliminates 
FT incorrect equations | AO1.1a | M1
Constructs a rigorous
mathematical argument to
show required result, clearly
constructing two correct
simultaneous equations and
eliminating AG | AO2.1 | R1
(b) | Solves a quadratic equation
and finds two values of  | AO3.1a | M1 | 3
r  ,r 6
2
1
r 6
2
3
r  8
2
828
so only one possible value of 
Finds two correct values of r | AO1.1b | B1
Finds both values of  | AO1.1b | A1
Gives a valid reason for
rejecting one of ‘their’ values | AO2.4 | R1
Total | 8
Q | Marking Instructions | AO | Marks | Typical Solution
The diagram shows a sector $AOB$ of a circle with centre $O$ and radius $r$ cm.

\includegraphics{figure_5}

The angle $AOB$ is $\theta$ radians

The sector has area 9 cm$^2$ and perimeter 15 cm.

\begin{enumerate}[label=(\alph*)]
\item Show that $r$ satisfies the equation $2r^2 - 15r + 18 = 0$
[4 marks]
\item Find the value of $\theta$. Explain why it is the only possible value.
[4 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 1  Q5 [8]}}
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