A pyramid has a square base with sides of length 4m and a height 3m. What is the length from one of the base corners to the top of the pyramid?

The student should first draw a diagram of the pyramid, labelling the lengths of the base and the height. The student is not able to solve this problem in one step but must think two steps ahead. The student must realise that in order to solve the question half the diagonal length of the base must be known. This can be found by using Pythagoras Theorem to find the hypotenuse of a right angle triangle with other sides being 2m. This half diagonal length is given by:l_B = ( (2m)² + (2m)² )^1/2 = ( 4m² + 4m⁴ )^1/2 = ( 8m² )^1/2 = (8)^1/2mHence this diagonal is the square root of 8 meters long. Knowing this a second right angle triangle should be drawn of base length sqrt(8) and height of 5 meters. Pythagoras should again be used to find the hypotenuse of this triangle:l = ( (sqrt(8)m)² + (3m)² )^1/2 = ( 8m² + 9m² )^1/2 = ( 8m² + 9m² )^1/2 = ( 17m² ) = sqrt(17)m