CAIE P1 2012 June — Question 3 5 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2012
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeCoplanar forces in equilibrium
DifficultyStandard +0.3 This is a straightforward geometry problem requiring calculation of areas using standard formulas. Students need to find the area of an equilateral triangle, subtract a circular sector, which involves recognizing the radius (√3 cm from the perpendicular from A to BC) and angle (120°). All steps are routine applications of memorized formulas with no novel problem-solving insight required.
Spec1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.05g Exact trigonometric values: for standard angles
3 \includegraphics[max width=\textwidth, alt={}, center]{4d8fcc3d-a2da-4d98-8500-075d10847be3-2_584_659_575_742} In the diagram, \(A B C\) is an equilateral triangle of side 2 cm . The mid-point of \(B C\) is \(Q\). An arc of a circle with centre \(A\) touches \(B C\) at \(Q\), and meets \(A B\) at \(P\) and \(A C\) at \(R\). Find the total area of the shaded regions, giving your answer in terms of \(\pi\) and \(\sqrt { } 3\).
Question 3:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(AQ\) (or \(r\)) \(= \sqrt{3}\)B1 soi. Allow 1.73
Area \(\Delta = \sqrt{3}\) (or area \(\Delta AQC = \frac{\sqrt{3}}{2}\))B1 [follow through] soi ft *their* \(\sqrt{3}\). Allow 1.73; ft *their* \(\sqrt{3}\). Allow 1.57. SCA1 for \(\pi/4\)
Area sector \(APR = \frac{1}{2}(\sqrt{3})^2 \times \frac{\pi}{3} = \frac{\pi}{2}\)M1A1 [follow through] from \(\frac{1}{2}(\sqrt{3})^2 \times \frac{\pi}{6}\) provided \(\Delta = \frac{\sqrt{3}}{2}\)
Shaded region \(= \sqrt{3} - \frac{\pi}{2}\) oe caoA1 [5] 
## Question 3:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $AQ$ (or $r$) $= \sqrt{3}$ | B1 | soi. Allow 1.73 |
| Area $\Delta = \sqrt{3}$ (or area $\Delta AQC = \frac{\sqrt{3}}{2}$) | B1 [follow through] | soi ft *their* $\sqrt{3}$. Allow 1.73; ft *their* $\sqrt{3}$. Allow 1.57. SCA1 for $\pi/4$ |
| Area sector $APR = \frac{1}{2}(\sqrt{3})^2 \times \frac{\pi}{3} = \frac{\pi}{2}$ | M1A1 [follow through] | from $\frac{1}{2}(\sqrt{3})^2 \times \frac{\pi}{6}$ provided $\Delta = \frac{\sqrt{3}}{2}$ |
| Shaded region $= \sqrt{3} - \frac{\pi}{2}$ oe cao | A1 [5] | |

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\includegraphics[max width=\textwidth, alt={}, center]{4d8fcc3d-a2da-4d98-8500-075d10847be3-2_584_659_575_742}

In the diagram, $A B C$ is an equilateral triangle of side 2 cm . The mid-point of $B C$ is $Q$. An arc of a circle with centre $A$ touches $B C$ at $Q$, and meets $A B$ at $P$ and $A C$ at $R$. Find the total area of the shaded regions, giving your answer in terms of $\pi$ and $\sqrt { } 3$.

\hfill \mbox{\textit{CAIE P1 2012 Q3 [5]}}
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