To quickly see the solution, one must view the little diagonal as the base of the triangle. Then draw a diagonal parallel to it through the large triangle. The perdendicular distances between these lines is the height of the triangle REGARDLESS of where you locate the "top" vertex. So, extend the vertical side of the middle square to this large diagonal. Now look at this new triangle with its base along the this extended line, length 4sqrt5, and height sqrt5. Thus the area equals 10!