1. A cylinder with radius 3.5 cm has its two ends closed.  If the total surface area is 209 cm², calculate the height of the cylinder.

(Take π  22)
               7

   (b)     In the diagram, O is the centre of the circle and ABC is a   tangent at B.  If BDF  =  66°  and  DBC  =  57°, calculate
         
                   (i)      EBF  and
                   (ii)     BGF.

 

observation

In the (a) part, they were required to find the height of a closed cylinder given the total surface area.  Thus from the formula; total surface area  A  =   2 πr(r + h) >> h  =  A/2πr  -  r   =     
   209 x 7      -   3.5   =  6.0 cm.
2 x 22 x 3.5

In the (b) part of the question, the candidates were required to determine the size of the given angles using circle theorems.  They were expected to show that: < EDB  =  90° (< on a semi circle).  <EDF  =  90 – 66  =  24°.  Thus, <EBF  =  24°  (< in the same segment as <EDF).  ÐBFD  =  57° (< in alternate segment to <DBC) 
   .
.   .  <BGF  =  180  -  (57 + 24)  =  99°.  This question was very unpopular among the candidates.