Browsing as Guest. Progress, bookmarks and attempts are disabled.
Log in to track your work.
Problem 191
191
The diagram shows a sector ABC of a circle with centre A and radius r. The line BD is perpendicular to AC. Angle CAB is θ radians.
(a) Given that θ = \frac{1}{6}π, find the exact area of BCD in terms of r.
(b) Given instead that the length of BD is \frac{\sqrt{3}}{2}r, find the exact perimeter of BCD in terms of r.
Solution
(a) The area of sector ABC is given by \(\frac{1}{2} r^2 \theta\). Substituting \(\theta = \frac{1}{6} \pi\), the sector area is \(\frac{1}{2} r^2 \times \frac{1}{6} \pi = \frac{1}{12} \pi r^2\).
The triangle ABD has base AD and height BD. Using trigonometry, \(BD = r \sin \frac{\pi}{6} = \frac{1}{2} r\) and \(AD = r \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} r\).
The area of triangle ABD is \(\frac{1}{2} \times \frac{1}{2} r \times \frac{\sqrt{3}}{2} r = \frac{\sqrt{3}}{8} r^2\).
Therefore, the area of BCD is \(\frac{1}{12} \pi r^2 - \frac{\sqrt{3}}{8} r^2\).
(b) Given BD = \frac{\sqrt{3}}{2} r, use trigonometry to find angle BAC: \(\angle BAC = \sin^{-1} \left( \frac{\sqrt{3}}{2} \right) = \frac{\pi}{3}\).
Then, \(AD = r \cos \frac{\pi}{3} = \frac{1}{2} r\) and \(CD = \frac{1}{2} r\).
The length of arc BC is \(r \times \frac{\pi}{3}\).
Thus, the perimeter of BCD is \(\frac{\sqrt{3}}{2} r + \frac{1}{2} r + \frac{\pi}{3} r\).
