CAIE P1 2018 November — Question 9 8 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2018
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAreas by integration
TypeArea of sector/segment problems
DifficultyChallenging +1.2 This is a multi-step geometry problem requiring calculation of lengths using trigonometry, then finding areas of two circular sectors and subtracting triangular areas. While it involves several steps and careful geometric reasoning, the techniques are all standard (right-angled trigonometry, sector area formula) with no novel insight required. The main challenge is careful bookkeeping across multiple steps, making it moderately above average difficulty.
Spec1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta

9 \includegraphics[max width=\textwidth, alt={}, center]{d178603a-f59a-4986-b5ab-b47eceedb2fc-14_465_677_260_733} The diagram shows a triangle \(O A B\) in which angle \(A B O\) is a right angle, angle \(A O B = \frac { 1 } { 5 } \pi\) radians and \(A B = 5 \mathrm {~cm}\). The arc \(B C\) is part of a circle with centre \(A\) and meets \(O A\) at \(C\). The arc \(C D\) is part of a circle with centre \(O\) and meets \(O B\) at \(D\). Find the area of the shaded region.

Question 9:
AnswerMarks Guidance
AnswerMarks Guidance
Angle \(OAB = \pi/2 - \pi/5 = 3\pi/10\)B1 Allow \(54°\) or \(0.9425\) rads
Sector \(CAB = \frac{1}{2} \times \left(\text{their } \frac{3\pi}{10}\right) \times 5^2\)M1 Expect \(11.78\)
\(OA = \dfrac{5}{\sin\frac{\pi}{5}} = 8.507\)M1A1 May be implied by \(OC = 3.507\)
Sector \(COD = \frac{1}{2} \times (\text{their } 3.507)^2 \times \frac{\pi}{5}\)M1 Expect \(3.86\)
\(\triangle OAB = \frac{1}{2} \times 5 \times (\text{their } 8.507)\sin\frac{3\pi}{10}\)M1 Or \(\frac{1}{2} \times 5 \times \dfrac{5}{\tan\frac{\pi}{5}}\) or \(2.5 \times \sqrt{(\text{their } 8.507)^2 - 25}\)
\(= 17.20\) or \(17.21\)A1
Shaded area \(17.20(\text{or } 17.21) - 11.78 - 3.86 = 1.56\) or \(1.57\)A1
8
## Question 9:

| Answer | Marks | Guidance |
|--------|-------|----------|
| Angle $OAB = \pi/2 - \pi/5 = 3\pi/10$ | **B1** | Allow $54°$ or $0.9425$ rads |
| Sector $CAB = \frac{1}{2} \times \left(\text{their } \frac{3\pi}{10}\right) \times 5^2$ | **M1** | Expect $11.78$ |
| $OA = \dfrac{5}{\sin\frac{\pi}{5}} = 8.507$ | **M1A1** | May be implied by $OC = 3.507$ |
| Sector $COD = \frac{1}{2} \times (\text{their } 3.507)^2 \times \frac{\pi}{5}$ | **M1** | Expect $3.86$ |
| $\triangle OAB = \frac{1}{2} \times 5 \times (\text{their } 8.507)\sin\frac{3\pi}{10}$ | **M1** | Or $\frac{1}{2} \times 5 \times \dfrac{5}{\tan\frac{\pi}{5}}$ or $2.5 \times \sqrt{(\text{their } 8.507)^2 - 25}$ |
| $= 17.20$ or $17.21$ | **A1** | |
| Shaded area $17.20(\text{or } 17.21) - 11.78 - 3.86 = 1.56$ or $1.57$ | **A1** | |
| | **8** | |

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9\\
\includegraphics[max width=\textwidth, alt={}, center]{d178603a-f59a-4986-b5ab-b47eceedb2fc-14_465_677_260_733}

The diagram shows a triangle $O A B$ in which angle $A B O$ is a right angle, angle $A O B = \frac { 1 } { 5 } \pi$ radians and $A B = 5 \mathrm {~cm}$. The arc $B C$ is part of a circle with centre $A$ and meets $O A$ at $C$. The arc $C D$ is part of a circle with centre $O$ and meets $O B$ at $D$. Find the area of the shaded region.\\

\hfill \mbox{\textit{CAIE P1 2018 Q9 [8]}}