CAIE P1 2018 June — Question 5 5 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2018
SessionJune
Marks5
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Mark schemeDownload PDF ↗
TopicAreas by integration
TypeArea of sector/segment problems
DifficultyModerate -0.5 This is a straightforward sector/segment problem requiring arc length to find the angle (θ = s/r = 6/5), then calculating sector area and subtracting triangle area. It involves standard formulas with no conceptual difficulty, though requires careful organization of multiple steps. Slightly easier than average due to being a routine application of well-practiced techniques.
Spec1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta
5 \includegraphics[max width=\textwidth, alt={}, center]{ea402a1d-3632-4637-9198-2365715b5246-06_323_775_260_685} The diagram shows a triangle \(O A B\) in which angle \(O A B = 90 ^ { \circ }\) and \(O A = 5 \mathrm {~cm}\). The arc \(A C\) is part of a circle with centre \(O\). The arc has length 6 cm and it meets \(O B\) at \(C\). Find the area of the shaded region.
Question 5:
AnswerMarks Guidance
Angle \(AOC = \frac{6}{5}\) or \(1.2\)M1 Allow 68.8°. Allow \(\frac{5}{6}\)
\(AB = 5 \times \tan(\textit{their } 1.2)\) OR by e.g. Sine Rule, Expect 12.86DM1 OR \(OB = \frac{5}{\cos(\textit{their } 1.2)}\). Expect 13.80
Area \(\triangle OAB = \frac{1}{2} \times 5 \times \textit{their } 12.86\), Expect 32.15DM1 OR \(\frac{1}{2} \times 5 \times \textit{their } OB \times \sin(\textit{their } 1.2)\)
Area sector \(\frac{1}{2} \times 5^2 \times \textit{their } 1.2\), Expect 15DM1 All DM marks are dependent on the first M1
Shaded region \(= 32.15 - 15 = 17.2\)A1 Allow degrees used appropriately throughout. 17.25 scores A0
Total: 5 
## Question 5:

| Angle $AOC = \frac{6}{5}$ or $1.2$ | M1 | Allow 68.8°. Allow $\frac{5}{6}$ |
|---|---|---|
| $AB = 5 \times \tan(\textit{their } 1.2)$ OR by e.g. Sine Rule, Expect 12.86 | DM1 | OR $OB = \frac{5}{\cos(\textit{their } 1.2)}$. Expect 13.80 |
| Area $\triangle OAB = \frac{1}{2} \times 5 \times \textit{their } 12.86$, Expect 32.15 | DM1 | OR $\frac{1}{2} \times 5 \times \textit{their } OB \times \sin(\textit{their } 1.2)$ |
| Area sector $\frac{1}{2} \times 5^2 \times \textit{their } 1.2$, Expect 15 | DM1 | All DM marks are dependent on the first M1 |
| Shaded region $= 32.15 - 15 = 17.2$ | A1 | Allow degrees used appropriately throughout. 17.25 scores A0 |
| **Total: 5** | | |

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5\\
\includegraphics[max width=\textwidth, alt={}, center]{ea402a1d-3632-4637-9198-2365715b5246-06_323_775_260_685}

The diagram shows a triangle $O A B$ in which angle $O A B = 90 ^ { \circ }$ and $O A = 5 \mathrm {~cm}$. The arc $A C$ is part of a circle with centre $O$. The arc has length 6 cm and it meets $O B$ at $C$. Find the area of the shaded region.\\

\hfill \mbox{\textit{CAIE P1 2018 Q5 [5]}}
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