CAIE P1 2015 June — Question 2 4 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2015
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRadians, Arc Length and Sector Area
TypeTriangle and sector combined - algebraic/general expressions
DifficultyModerate -0.8 This is a straightforward application of standard formulas for sector and semicircle areas. Students need to find the area of semicircle AYB and subtract the area of sector OAXB, using the given radius r and angle 2θ. The setup is clear, requires only direct formula application (sector area = ½r²(2θ), semicircle area = ½π(r/2)² after finding the semicircle's radius), and involves minimal algebraic manipulation—typical of early Pure 1 content with no problem-solving insight required.
Spec1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta

\includegraphics{figure_2} In the diagram, \(AYB\) is a semicircle with \(AB\) as diameter and \(OAXB\) is a sector of a circle with centre \(O\) and radius \(r\). Angle \(AOB = 2\theta\) radians. Find an expression, in terms of \(r\) and \(\theta\), for the area of the shaded region. [4]

Question 2:
AnswerMarks
2Radius of semicircle = 1AB = rsinθ
2
Area of semicircle = 1 2 πr²sin²θ = A 1
Shaded area = semicircle – segment
AnswerMarks
= A 1 − 1 2 r²2θ + 1 2 r²sin2θB1
B1
B1B1
AnswerMarks
[4]aef
Uses 1πr² with r = f(θ)
2
B1 ( –sector ), B1 for + (triangle)
Question 2:
2 | Radius of semicircle = 1AB = rsinθ
2
Area of semicircle = 1 2 πr²sin²θ = A 1
Shaded area = semicircle – segment
= A 1 − 1 2 r²2θ + 1 2 r²sin2θ | B1
B1
B1B1
[4] | aef
Uses 1πr² with r = f(θ)
2
B1 ( –sector ), B1 for + (triangle)
\includegraphics{figure_2}

In the diagram, $AYB$ is a semicircle with $AB$ as diameter and $OAXB$ is a sector of a circle with centre $O$ and radius $r$. Angle $AOB = 2\theta$ radians. Find an expression, in terms of $r$ and $\theta$, for the area of the shaded region. [4]

\hfill \mbox{\textit{CAIE P1 2015 Q2 [4]}}