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