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2011 WAEC Mathematics Theory A sector of a circle with radius 21 cm has an area of 280\(cm^{2}\). (a)...

Mathematics
WAEC 2011

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 \(\pi = \frac{22}{7}\)].

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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\(^o\).


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