Standard +0.3 This is a standard sector-segment problem requiring identification of geometric components (sector area minus triangle area, plus right triangle), application of basic formulas (A = ½r²θ, triangle area), and use of trigonometry (tan 60°). While it involves multiple steps and coordinate geometry, the approach is routine for A-level students who have covered circular measure, making it slightly easier than average.
3
In the diagram, \(A C\) is an arc of a circle, centre \(O\) and radius 6 cm . The line \(B C\) is perpendicular to \(O C\) and \(O A B\) is a straight line. Angle \(A O C = \frac { 1 } { 3 } \pi\) radians. Find the area of the shaded region, giving your answer in terms of \(\pi\) and \(\sqrt { } 3\).
Area of triangle as \(\frac{1}{2}bh\) or \(\frac{1}{2}ab\sin C\). Area of Sector. Co. Must be in this form, not decimals
\(\rightarrow 18\sqrt{3} - 6\pi\)
A1
[5] No \(\sqrt{3}\), max 3 out of 5
## Question 3:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\tan 60 = BC \div 6$ | M1 | Use of $\tan = \text{opp} \div \text{adj}$ |
| $BC = 6\sqrt{3}$ | A1 | In this form somewhere with $\sqrt{3}$ |
| Area $= \frac{1}{2} \times 6 \times \text{"BC"} - \frac{1}{2} \times 6^2 \times \frac{\pi}{3}$ | M1 M1 | Area of triangle as $\frac{1}{2}bh$ or $\frac{1}{2}ab\sin C$. Area of Sector. Co. Must be in this form, not decimals |
| $\rightarrow 18\sqrt{3} - 6\pi$ | A1 | [5] No $\sqrt{3}$, max 3 out of 5 |
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3
In the diagram, $A C$ is an arc of a circle, centre $O$ and radius 6 cm . The line $B C$ is perpendicular to $O C$ and $O A B$ is a straight line. Angle $A O C = \frac { 1 } { 3 } \pi$ radians. Find the area of the shaded region, giving your answer in terms of $\pi$ and $\sqrt { } 3$.
\hfill \mbox{\textit{CAIE P1 2004 Q3 [5]}}