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\(^o\).