A pyramid has a square base ABCD of side 40 cm.  The vertex V is 35 cm above the centre, O. of the base.  Calculate:

(a)        the angle VCO correct to the nearest degree;
(b)        |VC|;

• the area of a triangular face of the pyramid, correct to one decimal place.

observation

This question was not popular.  Only a few candidates attempted it.  Candidates found it pretty difficult.  Many of them could not draw the correct diagram and hence could not solve the problem.

From the diagram, |AC| =        √40² + 40² =   40√2   0r 56.56 cm
Therefore, OC = ½ AC = 20√2   or 28.28 cm.  <VCO = tan-1    35       = 51°                                                                                                    20√2  
to the nearest degree.  |VC| =         √ 35² + (20√2)²   =   45 cm

Taking ∆ VBC, height =       √VO² + (½ DC)²  =     √  35² + 20²  =  40.31 cm

Area of  ∆ VBC  =  20 cm x 40.31 cm   =  806.2 cm²