Question

12(a). A water reservoir in the form of a cone mounted on a hemisphere is built such that the plane face of the hemisphere fits exactly the base of the cone and the height of the cone is 6 times the radius of its base. (a) Illustrate this information in a diagram. (b) If the volume of the reservoir is 333⅓πm³, calculate, correct to the nearest whole number: (i) the volume of the hemisphere; (ii) the total surface area of the reservoir. [Take π=22/7]

Accepted Answer

(a) A diagram showing a cone of height h=6r sitting on top of a hemisphere of the same radius r, joined at their common circular base.
(b) Total volume = volume of cone + volume of hemisphere: 333⅓π = (1/3)πr²(6r) + (2/3)πr³ = 2πr³+(2/3)πr³ = (8/3)πr³. Solving: r³=125, r=5m.
(i) Volume of hemisphere = (2/3)πr³ = (2/3)×(22/7)×125 ≈ 262m³.
(ii) Slant height of cone l=√(r²+h²)=√(25+900)=√925≈30.4m. Total surface area = πrl+2πr² (curved surface of cone + curved surface of hemisphere) ≈ (22/7×5×30.4)+(2×22/7×25) ≈ 477.7+157.1 ≈ 634.8m².

WAEC Mathematics 2015 Theory — Question 21

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