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Problem 267
267
The diagram represents a metal plate OABC, consisting of a sector OAB of a circle with centre O and radius r, together with a triangle OCB which is right-angled at C. Angle AOB = \(\theta\) radians and OC is perpendicular to OA.
(i) Find an expression in terms of r and \(\theta\) for the perimeter of the plate.
(ii) For the case where r = 10 and \(\theta = \frac{1}{5}\pi\), find the area of the plate.
Solution
(i) The perimeter of the plate consists of the arc AB and the sides OC, CB, and OA. The length of the arc AB is given by \(r\theta\). The side OC is \(r \sin \theta\) and CB is \(r \cos \theta\). The side OA is r. Therefore, the perimeter is:
\(r\theta + r \sin \theta + r \cos \theta + r = r(1 + \theta + \cos \theta + \sin \theta)\)
(ii) The area of the plate consists of the area of the sector OAB and the area of the triangle OCB. The area of the sector OAB is \(\frac{1}{2} r^2 \theta\). For r = 10 and \(\theta = \frac{1}{5}\pi\), this becomes:
The area of the triangle OCB is \(\frac{1}{2} \times OC \times CB\). With OC = \(10 \sin \frac{\pi}{5}\) and CB = \(10 \cos \frac{\pi}{5}\), the area is:
