Observation

The Chief Examiner reported that this question was popular among candidates’, indicating good understanding of the question.
In part(a), a good percentage of candidates that attempted it could not apply the pythagoras theorems where necessary and in part(b), some candidates’ were able to apply the
trigonometry ratio correctly. However a good number of them could not get the trigonometry ratio correctly.
In part(a), they were expected to substitute what is given into the curved surface area formula to get 242 =  r (r + 4). Simplifying gives    r2 + 4r – 77 = 0. Factorizing gives (r – 7)(r + 11) = 0. Solving gives r = – 11, r = 7. Therefore,  r = 7.0 cm.
In part(b), they did as expected to recall that the radius, slant height and the height forms a right angled triangle. Using Pythagoras theorem gives h2 + r2 = (r + 4)2. Then substituting value of the radius gives    h2 + 72 = 112. Simplifying  gives h2 = 72. Taking square root of both sides gives h =  .

In part(c), they did as expected substitute the given parameter into the volume relation to obtain, Volume =  72 . Simplifying gives  435.576 cm3.
Therefore, volume = 435.6 cm3(1 d.p.)