A piece of paper TUVW has the precise shape of a square, with each side length being 20 units. The point M is the midpoint of side TU and N is the midpoint of TW.
The paper is now folded along the line MN such that the point T touches the paper. The point V is then folded over a line PQ parallel to MN such that V lies on MN.
Determine the area of the hexagon NMUPQW.
Geometry is not my strong suit but this one seems sufficiently simple, so here goes:
The first fold removes triangle NTM from the square. This is an isosceles right triangle with base (MN) = 10sqrt(2) (half the length of the diagonal of the square), and height 5sqrt(2) (the height is half the length of the base). The area of this triangle is bh/2 = 50.
The second fold removes triangle PVQ from the square. This is also an isosceles right triangle with height 7.5sqrt(2) and base 15sqrt(2). The area of this triangle is bh/2 = 112.5.
The area of the original square was 400, so after removing these two triangles, the area of the hexagon that remains is 237.5.
Posted by tomarken
on 2014-05-06 10:12:47