Note:
For a P5 pupil who is solving this problem, so sorry, but you are forced to ASSUME that the shaded triangle at the bottom is an equilateral triangle, else it can't be solved. With that ASSUMPTION, one can then arrive at the final answer that the entire shaded area is 88cm2.
But this is so sad, as we should NEVER have to make such dangerous assumptions. Pupils who are very careful will be very hesitant to make this assumption and hence will waste precious time trying to solve this problem.
But this is so sad, as we should NEVER have to make such dangerous assumptions. Pupils who are very careful will be very hesitant to make this assumption and hence will waste precious time trying to solve this problem.
For a secondary school pupil, no assumption is needed as you can prove, by application of Pythagoras Theorem and then via algebraic manipulation, that the base of the isosceles triangle at the bottom is approx 4.04cm, which is about 4, or a third of 12 . With this finding, it goes to prove that the assumption is needed for the poor P5 kid, which again I am so sorry for those P5 pupils who had to sit for this question under a real exam setting! ><
** Look out for another question that I will be crafting to prove that the paper can be similarly folded yet not achieving a near equilateral triangle at the bottom! ><
