Whats the Shaded Area?

Let AB=BC= 2 units

1-> Area of triangle ABE= 1/2 AE BE sin(X)= 1/2* 2 * 1 sin (X) = sin(X) sq. units 2-> Area of triangle FEC= 1/2 FC CE sin(180-X)= 1/2* 1 * 1 sin (X) = 1/2 sin(X) sq. units 3-> Area of triangle ADF= 1/2 AD DF sin(X)= 1/2* 2 * 1 sin (X) = sin(X) sq. units

4->Area of the parallelogram= product of the sides * sine of the angle b/w them =4 sin(X) sq. units

Area of the shaded region= total area- sum of areas of the triangles = (4-1-1-1/2) sin(X) sq.units =3/2 sin(X) sq. units

Area of the shaded region/Area of the parallelogram: = (3/2)/4= 3/8

Given we are just dealing with a ratio of areas, we can just deform the figure into a rectangle and compute.

The Shaded Region becomes the Area of the "Rectangle" less the area of the three surrounding "Right Triangles". $\\ W*H - \frac{1}{2}(\frac{W}{2})(H) - \frac{1}{2}(W)(\frac{H}{2}) - \frac{1}{2}(\frac{W}{2})(\frac{H}{2}) = (W)(H)(1-\frac{1}{4}-\frac{1}{4}-\frac{1}{8}) \\$ Shaded Fraction = $\frac{(W)(H)(\frac{3}{8})}{(W)(H)} = \boxed{\frac{3}{8}}$