Standard +0.3 This is a straightforward geometry problem requiring identification of a right triangle, use of Pythagoras to find angle/sides, calculation of a sector area, and subtraction to find a segment area. All steps are standard techniques with no novel insight required, making it slightly easier than average.
4
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In the diagram, \(D\) lies on the side \(A B\) of triangle \(A B C\) and \(C D\) is an arc of a circle with centre \(A\) and radius 2 cm . The line \(B C\) is of length \(2 \sqrt { } 3 \mathrm {~cm}\) and is perpendicular to \(A C\). Find the area of the shaded region \(B D C\), giving your answer in terms of \(\pi\) and \(\sqrt { } 3\).
area \(\Delta = 2\sqrt{3}\); \(\tan A = \frac{2\sqrt{3}}{2} \Rightarrow A = \frac{\pi}{3}\); Area sector \(= \frac{1}{2} \times 2^2 \times \frac{\pi}{3} = \frac{2\pi}{3}\); Shaded area \(= 2\sqrt{3} - \frac{2\pi}{3}\)
B1, B1, M1, A1
Accept 60°; Use of \(\frac{1}{2}r^2\theta\) with \(\theta\) in radians; cao [4]
area $\Delta = 2\sqrt{3}$; $\tan A = \frac{2\sqrt{3}}{2} \Rightarrow A = \frac{\pi}{3}$; Area sector $= \frac{1}{2} \times 2^2 \times \frac{\pi}{3} = \frac{2\pi}{3}$; Shaded area $= 2\sqrt{3} - \frac{2\pi}{3}$ | B1, B1, M1, A1 | Accept 60°; Use of $\frac{1}{2}r^2\theta$ with $\theta$ in radians; cao [4]
4\\
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In the diagram, $D$ lies on the side $A B$ of triangle $A B C$ and $C D$ is an arc of a circle with centre $A$ and radius 2 cm . The line $B C$ is of length $2 \sqrt { } 3 \mathrm {~cm}$ and is perpendicular to $A C$. Find the area of the shaded region $B D C$, giving your answer in terms of $\pi$ and $\sqrt { } 3$.
\hfill \mbox{\textit{CAIE P1 2012 Q4 [4]}}