(a) Calculate the area of the shaded segment of the circle shown in the diagram...

MATHEMATICS
WAEC 1992

(a)

Calculate the area of the shaded segment of the circle shown in the diagram [Take \(\pi = \frac{22}{7}\)]

(b) A tin has radius 3cm and height 6cm. Find the (i) total surface area of the tin ; (ii) volume, in litres, that will fill the tin to capacity, correct to two decimal places.

[Take \(\pi = \frac{22}{7}\)]

Explanation

(a) Area of the shaded segment = Area of sector - Area of triangle

= \(\frac{\theta}{360°} \times \pi r^{2} - \frac{1}{2} ab \sin \theta\)

Area of sector = \(\frac{63}{360} \times \frac{22}{7} \times 10^{2}\)

= \(55 cm^{2}\)

Area of triangle = \(\frac{1}{2} \times 10 \times 10 \times \sin 63°\)

= \(50 \times 0.891 = 44.55 cm^{2}\)

\(\therefore \text{Area of the shaded segment} = 55 - 44.55 = 10.45 cm^{2}\)

(b) (i) Total surface area = \(2\pi r (r + h)\)

= \(2 \times \frac{22}{7} \times 3 (3 + 6)\)

= \(\frac{132}{7} \times 9 = \frac{1188}{7}\)

= \(169.71 cm^{2}\)

(ii) Volume of tin = \(\pi r^{2} h\)

= \(\frac{22}{7} \times 3 \times 3 \times 6 = \frac{1188}{7}\)

\(\approxeq 169.71 cm^{3}\)



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