A square based pyramid with corners ABCD has side length 6 cm. The distance from the centre of the square (C) to the top vertex of the pyramid (V) is 4 cm. Work out the total surface area of the pyramid.

The first thing I would do with this question is to draw it out to help visualise the shape.From this drawing you can see that the shape has 4 triangle shaped faces (let's call this triangle T1) and 1 square face for its surface area. We know the 4 triangle faces are all the same area as the question says the pyramid is square based. Now you just need to find the area of the square and of the triangle.The area of the square is just 66=36 cm^2 as we know the side length of the square is 6cm.The area of the triangle (T1) is a little more difficult. This is worked out by multiplying the length of the triangle by the height of the triangle, then dividing this answer by 2. We know the length of the triangle is 6cm as this is also the side length of the square base. To find out the height we need to work out the distance between the midpoint of the side of the square (M) and the point V. This can be worked out using Pythagoras' theorem on another triangle (let's call this triangle T2), where one edge is the distance from C to V, one edge is the distance from C to M and the hypotenuse (long edge) is the distance from M to V. Pythagoras' theorem is a^2 +b^2 = c^2. Here, a=3 (as this is half of 6) b=4 and c is unknown. Using this we can find that c=5 and so the height of the triangle is 5. Therefore, the area of the triangle is (56)/2= 15 cm^2Now to work out the total surface area we do 4*(15) + 36 = 96 cm^2