CAIE P1 2019 November — Question 4 5 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2019
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAreas by integration
TypeArea of sector/segment problems
DifficultyStandard +0.3 This is a standard geometric problem requiring identification of an equilateral triangle (angle CAO = π/3), then applying sector and triangle area formulas. The algebra is straightforward with minimal steps, making it slightly easier than average for A-level.
Spec1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta

4 \includegraphics[max width=\textwidth, alt={}, center]{17e813c6-890f-4198-b20a-557b133e8c34-05_360_639_255_753} The diagram shows a semicircle \(A C B\) with centre \(O\) and radius \(r\). Arc \(O C\) is part of a circle with centre \(A\).
  1. Express angle \(C A O\) in radians in terms of \(\pi\).
  2. Find the area of the shaded region in terms of \(r , \pi\) and \(\sqrt { } 3\), simplifying your answer.

Question 4:
Part (i)
AnswerMarks
Angle \(CAO = \frac{\pi}{3}\)B1
Part (ii)
AnswerMarks Guidance
\((\text{Sector } AOC) = \frac{1}{2}r^2 \times their\,\frac{\pi}{3}\)M1 SOI
\((\Delta ABC) = \frac{1}{2}(r)(2r)\sin\!\left(their\,\frac{\pi}{3}\right)\) or \(\frac{1}{2}(2r)(r)\frac{\sqrt{3}}{2}\) or \(\frac{1}{2}(r)(r)\sqrt{3}\)M1 For M1M1, \(their\,\frac{\pi}{3}\) must be of the form \(k\pi\) where \(0 < k < \frac{1}{2}\)
\((\Delta ABC) = \frac{1}{2}(r)(2r)\sin\!\left(\frac{\pi}{3}\right)\) or \(\frac{1}{2}(2r)(r)\frac{\sqrt{3}}{2}\) or \(\frac{1}{2}(r)(r)\sqrt{3}\)A1 All correct
\(r^2\!\left(\frac{\sqrt{3}}{2}\right) - \frac{1}{2}r^2\!\left(\frac{\pi}{3}\right)\)A1
## Question 4:

**Part (i)**

Angle $CAO = \frac{\pi}{3}$ | **B1** |

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**Part (ii)**

$(\text{Sector } AOC) = \frac{1}{2}r^2 \times their\,\frac{\pi}{3}$ | **M1** | SOI

$(\Delta ABC) = \frac{1}{2}(r)(2r)\sin\!\left(their\,\frac{\pi}{3}\right)$ **or** $\frac{1}{2}(2r)(r)\frac{\sqrt{3}}{2}$ **or** $\frac{1}{2}(r)(r)\sqrt{3}$ | **M1** | For M1M1, $their\,\frac{\pi}{3}$ must be of the form $k\pi$ where $0 < k < \frac{1}{2}$

$(\Delta ABC) = \frac{1}{2}(r)(2r)\sin\!\left(\frac{\pi}{3}\right)$ **or** $\frac{1}{2}(2r)(r)\frac{\sqrt{3}}{2}$ **or** $\frac{1}{2}(r)(r)\sqrt{3}$ | **A1** | All correct

$r^2\!\left(\frac{\sqrt{3}}{2}\right) - \frac{1}{2}r^2\!\left(\frac{\pi}{3}\right)$ | **A1** |
4\\
\includegraphics[max width=\textwidth, alt={}, center]{17e813c6-890f-4198-b20a-557b133e8c34-05_360_639_255_753}

The diagram shows a semicircle $A C B$ with centre $O$ and radius $r$. Arc $O C$ is part of a circle with centre $A$.\\
(i) Express angle $C A O$ in radians in terms of $\pi$.\\

(ii) Find the area of the shaded region in terms of $r , \pi$ and $\sqrt { } 3$, simplifying your answer.\\

\hfill \mbox{\textit{CAIE P1 2019 Q4 [5]}}