A sector of a circle with radius 21 cm has an area of 280\(cm^{2}\). (a) Calculate, correct to 1 decimal place, the perimeter of the sector. (b) If the sector is bent such that its straight edges coincide to form a cone, calculate, correct to the nearest degree, the vertical angle of the cone. [Take \(\pi = \frac{22}{7}\)].
Explanation
(a) Area of sector = \(\frac{\theta}{360} \times \pi r^{2}\) \(280 = \frac{\theta}{360} \times \frac{22}{7} \times 21 \times 21\) \(280 = \frac{1386 \theta}{360}\) \(\theta = \frac{280 \times 360}{1386}\) \(\theta = 72.72°\) Perimeter of sector = \(2r + \frac{\theta}{360} \times 2\pi r\) = \(2(21) + \frac{72.72}{360} \times 2 \times \frac{22}{7} \times 21\) = \(42 + (0.202 \times 2 \times 66)\) = \(42 + 26.667\) = \(68.667 cm\) \(\approxeq 68.7 cm\) (b) When the sector is bent to form a cone, its radius becomes the slant height of the cone. The radius of the base of the cone is obtained from the relation \(r = \frac{R \theta}{360}\), where r = radius of the base of the cone, R = radius of the sector, θ = angle of the sector. Therefore, r = \(\frac{21 \times 800}{11 \times 360}\) = \(\frac{140}{33}\) If y is the vertical angle of the cone, then \(sin \frac{y}{2} = \frac{r}{l}\) = \(\frac{140}{33 \times 21}\) = 0.2020 Hence, required angle = y = 2 x sin\(^{-1}\) (0.2020) = 23\(^o\).
