Perfect Polygons

The red area is the area of equilateral triangle $ACE$ minus the area of right-triangle $ABC$ .

$\triangle ABC$ is a $45^\circ - 45^\circ - 90^\circ$ right-triangle with hypotenuse of length $2,$ so it has legs of length $\frac{2}{\sqrt{2}}.$ Thus, its area is $\frac{1}{2} \cdot \frac{2}{\sqrt 2} \cdot \frac{2}{\sqrt 2} = 1.$

On the other hand, $\triangle ACE$ has sides of length $2,$ so its area is $\frac{2^2 \sqrt 3}{4} = \sqrt{3}.$

Thus, the area of the shaded region is $\sqrt{3}-1.$

By pythagorean theorem, we have

$2^2=x^2+x^2$ $\implies$ $x^2=2$

The area of the quadrilateral $AECB$ is equal to the area of equilateral triangle $CAE$ minus the area of triangle $CAB$ . We have

$A_{AECB}=\dfrac{\sqrt{3}}{4}(2^2)-\dfrac{1}{2}(2)=$ $\boxed{\sqrt{3}-1}$

if the diagonals intersect at O

area of COB = 1/2

and area of COB =1/2 *SQR (3)

area of CBE = 1/2 *SQR (3) - 1/2

area of quadrilateral = 2 * ( 1/2 *SQR (3) - 1/2)

= SQR (3) - 1