CAIE P1 2024 November — Question 7 10 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2024
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRadians, Arc Length and Sector Area
TypeCompound shape area
DifficultyStandard +0.3 This is a multi-part question requiring standard applications of sector area, triangle area, and semicircle formulas. Part (a) involves straightforward substitution into formulas with given angle. Part (b) requires working backwards from semicircle area to find radius, then calculating perimeter components. While it has multiple steps and requires careful organization of five different regions, all techniques are routine for P1 level with no novel problem-solving insight needed.
Spec1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta2.02f Measures of average and spread
\includegraphics{figure_7} The diagram shows a metal plate \(ABCDEF\) consisting of five parts. The parts \(BCD\) and \(DEF\) are semicircles. The part \(BAFO\) is a sector of a circle with centre \(O\) and radius 20 cm, and \(D\) lies on this circle. The parts \(OBD\) and \(ODF\) are triangles. Angles \(BOD\) and \(DOF\) are both \(\theta\) radians.
  1. Given that \(\theta = 1.2\), find the area of the metal plate. Give your answer correct to 3 significant figures. [5]
  2. Given instead that the area of each semicircle is \(50\pi \text{ cm}^2\), find the exact perimeter of the metal plate. [5]
Question 7:
AnswerMarks
7(a)1
Area of sector BOF = 202(2π−2.4) =776.63
AnswerMarks Guidance
2M1 Or combination of large semi-circle and small sector:
1 1
202π+ 202(π−2.4).
2 2
Length BD = DF = 220sin0.6 or 202+202−22020cos1.2
AnswerMarks Guidance
=22.58M1* Length of radius of small circles is acceptable for M1.
π(20sin0.6)2 =400.64
AnswerMarks
Area of two semicircles =DM1
1
Area of triangles = 2 2020sin1.2 =372.81
AnswerMarks Guidance
2M1
Total area = 1550 [cm2 ]A1 Expect1550.09but accept AWRT to 3sf.
5
AnswerMarks
7(b)1
πr2 =50π  r=10
AnswerMarks Guidance
2B1  
May be seen as 20sin , where = .
2 3
π
⇒ =
AnswerMarks Guidance
3M1* OE
Findingusing their r. Allow working in degrees.
 2π
Arc length of sector BOF = 202−their 
AnswerMarks
 3 DM1
 2π

Total perimeter = 202π−their +2πtheir10

AnswerMarks Guidance
 3 DM1 Dependent on the first dM1.
140π 2
or 46 π
AnswerMarks Guidance
3 3A1 Must be a single exact term.
5
AnswerMarks Guidance
QuestionAnswer Marks 
Question 7:
--- 7(a) ---
7(a) | 1
Area of sector BOF = 202(2π−2.4) =776.63
2 | M1 | Or combination of large semi-circle and small sector:
1 1
202π+ 202(π−2.4).
2 2
Length BD = DF = 220sin0.6 or 202+202−22020cos1.2
=22.58 | M1* | Length of radius of small circles is acceptable for M1.
π(20sin0.6)2 =400.64
Area of two semicircles = | DM1
1
Area of triangles = 2 2020sin1.2 =372.81
2 | M1
Total area = 1550 [cm2 ] | A1 | Expect1550.09but accept AWRT to 3sf.
5
--- 7(b) ---
7(b) | 1
πr2 =50π  r=10
2 | B1 |  
May be seen as 20sin , where = .
2 3
π
⇒ =
3 | M1* | OE
Findingusing their r. Allow working in degrees.
 2π
Arc length of sector BOF = 202−their 
 3  | DM1
 2π
Total perimeter = 202π−their +2πtheir10
 3  | DM1 | Dependent on the first dM1.
140π 2
or 46 π
3 3 | A1 | Must be a single exact term.
5
Question | Answer | Marks | Guidance
\includegraphics{figure_7}

The diagram shows a metal plate $ABCDEF$ consisting of five parts. The parts $BCD$ and $DEF$ are semicircles. The part $BAFO$ is a sector of a circle with centre $O$ and radius 20 cm, and $D$ lies on this circle. The parts $OBD$ and $ODF$ are triangles. Angles $BOD$ and $DOF$ are both $\theta$ radians.

\begin{enumerate}[label=(\alph*)]
\item Given that $\theta = 1.2$, find the area of the metal plate. Give your answer correct to 3 significant figures. [5]

\item Given instead that the area of each semicircle is $50\pi \text{ cm}^2$, find the exact perimeter of the metal plate. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2024 Q7 [10]}}
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