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 2pi 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).