Browsing as Guest. Progress, bookmarks and attempts are disabled.
Log in to track your work.
9709 P13 - Nov 2014 - Q2
245
In the diagram, OADC is a sector of a circle with centre O and radius 3 cm. AB and CB are tangents to the circle and angle ABC = \(\frac{1}{3} \pi\) radians. Find, giving your answer in terms of \(\sqrt{3}\) and \(\pi\),
(i) the perimeter of the shaded region,
(ii) the area of the shaded region.
Solution
(i) To find the perimeter of the shaded region, we need to calculate the lengths of AB, BC, and the arc AC.
The length of AB or CB is given by \(\frac{3}{\tan \frac{\pi}{6}} = 3 \tan \frac{\pi}{3} = 3\sqrt{3}\).
The length of the arc AC is \(3 \times \frac{2\pi}{3} = 2\pi\).
Therefore, the perimeter of the shaded region is \(3\sqrt{3} + 3\sqrt{3} + 2\pi = 6\sqrt{3} + 2\pi\).
(ii) To find the area of the shaded region, we calculate the area of triangle OABC and subtract the area of sector OADC.
The area of triangle OABC is \(2 \times \frac{1}{2} \times 3 \times 3\sqrt{3} = 9\sqrt{3}\).
The area of sector OADC is \(\frac{1}{2} \times 3^2 \times \frac{2\pi}{3} = 3\pi\).
Thus, the area of the shaded region is \(9\sqrt{3} - 3\pi\).
