CAIE P1 2024 June — Question 7 8 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2024
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRadians, Arc Length and Sector Area
TypeShaded region between arcs
DifficultyStandard +0.3 This is a straightforward application of arc length formulas and basic trigonometry. Part (a) requires using the parallel lines condition to find θ (likely via perpendicular distances), which is routine geometry. Part (b) applies standard arc length and sector area formulas with minimal problem-solving required. The multi-step nature and 8 total marks place it slightly above average, but all techniques are standard P1 content with no novel insight needed.
Spec1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta
\includegraphics{figure_7} In the diagram, \(AOD\) and \(BC\) are two parallel straight lines. Arc \(AB\) is part of a circle with centre \(O\) and radius \(15\text{cm}\). Angle \(BOA = \theta\) radians. Arc \(CD\) is part of a circle with centre \(O\) and radius \(10\text{cm}\). Angle \(COD = \frac{1}{3}\pi\) radians.
  1. Show that \(\theta = 0.7297\), correct to 4 decimal places. [1]
  2. Find the perimeter and the area of the shape \(ABCD\). Give your answers correct to 3 significant figures. [7]
Question 7:
AnswerMarks
7(a)π 10 10
Angle  = cos1 or sin1 0.7297
AnswerMarks Guidance
2 15 15B1 Condone working in degrees if converted to radians at the
end.
AG
1
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
7(b)BC = 152 102 11.18... or 55
 B1
Arc AB = 150.7297 10.9455B1
Perimeter = their BC + their arc AB + 25 + 5πM1
Perimeter = 62.8A1 AWRT
1
Area sector AOB = 1520.7297 82.09
AnswerMarks
2B1
1 π
Area = 10their BC + their sector AOB + 102
AnswerMarks Guidance
2 4M1
Area = 217A1 AWRT
7
AnswerMarks Guidance
QuestionAnswer Marks 
Question 7:
--- 7(a) ---
7(a) | π 10 10
Angle  = cos1 or sin1 0.7297
2 15 15 | B1 | Condone working in degrees if converted to radians at the
end.
AG
1
Question | Answer | Marks | Guidance
--- 7(b) ---
7(b) | BC = 152 102 11.18... or 55
  | B1
Arc AB = 150.7297 10.9455 | B1
Perimeter = their BC + their arc AB + 25 + 5π | M1
Perimeter = 62.8 | A1 | AWRT
1
Area sector AOB = 1520.7297 82.09
2 | B1
1 π
Area = 10their BC + their sector AOB + 102
2 4 | M1
Area = 217 | A1 | AWRT
7
Question | Answer | Marks | Guidance
\includegraphics{figure_7}

In the diagram, $AOD$ and $BC$ are two parallel straight lines. Arc $AB$ is part of a circle with centre $O$ and radius $15\text{cm}$. Angle $BOA = \theta$ radians. Arc $CD$ is part of a circle with centre $O$ and radius $10\text{cm}$. Angle $COD = \frac{1}{3}\pi$ radians.

\begin{enumerate}[label=(\alph*)]
\item Show that $\theta = 0.7297$, correct to 4 decimal places. [1]

\item Find the perimeter and the area of the shape $ABCD$. Give your answers correct to 3 significant figures. [7]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2024 Q7 [8]}}
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