**Question 12**

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 to the base of the cone and the height of the cone is 6 times the radius of its base.

- Illustrate this information in a diagram.
- If the volume of the reservoir is 333p m3, calculate, correct to the nearest whole number, the:

(i) volume of the hemisphere;

(ii) total surface area of the reservoir.

[Take p= ]

### Observation

The Chief Examiner reported that this question was quite unpopular among the candidates and the performance of those who attempted it was poor. Majority of the candidates were not able to represent the given information in a correct diagram. Others could not apply the formula correctly.

In part (a), candidates were expected to draw the following diagram.

In part (b), they were expected to show that the volume of the reservoir = volume of cone + volume of hemisphere. i.e. 333 = 2r3 + r3,

where r = radius of base of cone = radius of hemisphere. Solving this equation gave r = 5 m. Volume of hemisphere = r3 = 53 = 262 cm3. To get the total surface area of the reservoir, candidates were expected to obtain the slant height of the conical part which would lead them to obtain the curved surface area of the cone. The slant height, l = = = 30.4138 m. Total surface area of reservoir = curved surface area of cone + curved surface area of the hemisphere = rl + 2r2 = () + (2) = 635 m2, correct to the nearest whole number.