The diagram shows a sector $$\(C A B\)$$ which is part of a circle with centre $$\(C\)$$. A circle with centre $$\(O\)$$ and radius $$\(r\)$$ lies within the sector and touches it at $$\(D, E\)$$ and $$\(F\)$$, where $$\(C O D\)$$ is a straight line and angle $$\(A C D\)$$ is $$\(\theta\)$$ radians. It is now given that $$\(r=4\)$$ and $$\(\theta=\frac{1}{6} \pi\)$$. Find the area of the shaded region in terms of $$\(\pi\)$$ and $$\(\sqrt{3}\)$$. .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... ....................................................................................................................................................
Exam No：9709_w20_qp_11 Year：2020 Question No：10(c)
Answer:
Area $F O C=\frac{1}{2} \times 4 \times$ their $O C \times \sin \frac{\pi}{3}$
$8 \sqrt{3}$
Area sector $F O E=\frac{1}{2} \times \frac{2 \pi}{3} \times 4^{2}=\frac{16 \pi}{3}$
Shaded area $=16 \sqrt{3}-\frac{16 \pi}{3}$
Alternative method for question 10 (c)
$F C=\sqrt{(\text { their } O C)^{2}-4^{2}}$
Area $F O C=\frac{1}{2} \times 4 \times 4 \sqrt{3}=8 \sqrt{3}$
Area of half sector $F O E=\frac{1}{2} \times \frac{\pi}{3} \times 4^{2}=\frac{8 \pi}{3}$
Shaded area $=16 \sqrt{3}-\frac{16 \pi}{3}$
$8 \sqrt{3}$
Area sector $F O E=\frac{1}{2} \times \frac{2 \pi}{3} \times 4^{2}=\frac{16 \pi}{3}$
Shaded area $=16 \sqrt{3}-\frac{16 \pi}{3}$
Alternative method for question 10 (c)
$F C=\sqrt{(\text { their } O C)^{2}-4^{2}}$
Area $F O C=\frac{1}{2} \times 4 \times 4 \sqrt{3}=8 \sqrt{3}$
Area of half sector $F O E=\frac{1}{2} \times \frac{\pi}{3} \times 4^{2}=\frac{8 \pi}{3}$
Shaded area $=16 \sqrt{3}-\frac{16 \pi}{3}$
Knowledge points:
1.4.2 use the formulae in solving problems concerning the arc length and sector area of a circle (Including calculation of lengths and angles in triangles and areas of triangles.)
Solution:
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