Given a quadrilateral with vertices $ABCD$ clockwisely, let $O$ be the midpoint of the line segment $AD$. Suppose $\angle AOB = \angle BOC = \angle COD = \pi/3$ and $AB = a, CD = b, BC = \sqrt{a^2+b^2}$, where $a, b$ are two positive real numbers. Compute the area of this quadrilateral in terms of $a$ and $b$.
I have tried to use the law of cosines to formulate 3 equations:
$$\begin{cases} x^2+y^2-xy = a^2, \\ x^2+z^2-xz = b^2, \\ y^2+z^2-yz = a^2+b^2, \end{cases}$$ where $OA = OD = x$, $OB = y$, $OC = z$. Certainly I can solve these equations, however I would like to know if there is a nice geometric approach to obtain this area. Thanks!
In the simplest case, you have an inverted equilateral triangle sided by two half-equilateral triangles making it a rectangle and $BC\ne\sqrt{a^2+b^2}$. Sides a and b are $\sqrt{1-(\frac {1}{2})^2}= \frac{\sqrt{3}}{2}$ times the sides of an equilateral triangle side which is also the width of the "rectangle". This means the top and botom are the inverse ratio times a or b. Assuming $a=b=1$, the area is $h\times w=1\times\frac{2}{\sqrt{3}}$.
Making $BC=\sqrt{a^2+b^2}$ requires that they have lengths such as $3$ and $4$, respectively or $5,12$ or $7,24$ for an infinite number os solutions. (not true exactly) From here, assuming the left side where we have a triangle with sides $(a,\sqrt{a^2+b^2})$ and the right where we have a triangle with sides $(\sqrt{a^2+b^2},b)$, if we pulled these two together, the peak forms a right triangle having sides a and b. The bottom is the same as BC+ another BC and the altitude is the same as though there were two half-equilateral triangle beside the central inverted equilateral triangle.
This means the area is $$h\times width_{avg}=\frac{\sqrt{3}\sqrt{a^2+b^2}}{2}\cdot\frac{(2\sqrt{a^2+b^2}+\sqrt{a^2+b^2})}{2} =\frac{3\sqrt{3}(a^2+b^2)}{4}$$