Question 29865
If I get the picture correctly, you are trying to create a rhomb(sic) (in the U.S. this is called a rhombus) by skewing a 400 X 400-unit square so that the interior angles of the rhombus are: 72, 108, 72, and 108 degrees.

Let's assume that your initial square is based on the x-axis and the left
side is coincident with the y-axis, so the bottom-left corner is located at the origin of the coordiante system.

Now you'll displace the top of the square some distance to the right in the x-direction, leaving the bottom fixed of course. Your question is: What should the displacement distance be to obtain the required rhombus?

After the displacement, if you drop a perpendicular line from the top-left corner of the newly-formed rhombus to the base, you will have formed a right-triangle whose hypotenuse (h) is the length of the original square (400 units) and whose base angle is 72 degrees.

You can find the displacement distance (d) using the cosine of 72 degrees. Recall that the cosine of the angle of interest is:
Cosine = (side adjacent to the angle)/(the hypotenuse). The side adjacent is the displacent (d). So: 

{{{cos(72) = d/h}}} But h = 400
{{{cos(72) = d/400}}} Solve for d.
{{{d = 400cos(72)}}}
{{{d = 123.606}}}

The required displacement is 123.6 units (rounded to the nearest tenth)