Irregular exagon

By changing the position of the triangles you get to something like the photo above. Then the area labeled as A is just the area of a rectangular trapezium less the area of white triangles.

The white triangle on the left has a hypotenuse equal 5. So the left side of bottom white triangle is also 5 (since the left green triangle is equilateral). White triangle on the right has left leg equal 3 (since the top green triangle is equilateral), and the right leg also (the $45^\circ$ -angle makes it an isosceles triangle). So the hypotenuse of the right white triangle, and therefore also the right side of the bottom white triangle, is $3\times \sqrt{2}$ . The angle between these two known sides of the bottom white triangle is $360^\circ-3\times 60^\circ-45^\circ-arctan(\frac{4}{3})\approx 81.87^\circ$ . The area $A$ , calculated from two sides and an angle in between, is $A=\frac{1}{2}\times 5 \times 3\sqrt{2}\times sin(81.87^\circ)=10.5$

$x+y+45+60+60+60=360$

$y = 135 - x$

Law of Cosine in Triangle A.

$M^2 = (5)^2 + (3\sqrt{2})^2 - 2(5)(3\sqrt{2})\cos(y)$

$M^2 = 25 + 18 - (30\sqrt{2})\cos(135-x)$

$\Rightarrow\cos(135-x)=\cos(135)\cos(x)+\sin(135)\sin(x)$

$\Rightarrow\cos(135-x)=\displaystyle\frac{-\sqrt{2}}{2}\left(\frac{3}{5}\right)+\displaystyle\frac{\sqrt{2}}{2}\left(\frac{4}{5}\right)$

$\Rightarrow\cos(135-x)=\displaystyle\frac{\sqrt{2}}{10}$

$\displaystyle M^2 = 25 + 18 - (30\sqrt{2})\left(\frac{\sqrt{2}}{10}\right)$

$M^2 = 43 - 6$

$M = \sqrt{37} \text{ cm}$

Sides of Triangle A: $\displaystyle5\text{ cm}, 3\sqrt{2}\text{ cm}, \text{ and }\sqrt{37}\text{ cm}$

Heron's Formula for area of triangle given three sides:

$\displaystyle A=\sqrt{p(p-a)(p-b)(p-c)}$

where $\displaystyle{p=\frac{a+b+c}{2}}$

Solving for the area, we get $\boxed{\displaystyle\text{A }=10.5\text{ cm}}$