geometry
I know there is already a question about resolving a quadrilateral from three sides and two angles, but I want to ask about a special case. Firstly, two of the sides are known to be of equal size. Secondly, I'm only interested in the area, not in the remaining angles or lengths. Can anyone suggest a simple formula?
The wikipedia article specifies that the polygon is convex in almost every paragraph of this article.
It is only true for a convex polygon, you just proved that wikipedia can be wrong :).
For a convex polygon i can think of the following proof :
With 4 lengths and 1 diagonal you define 2 unique triangles (a triangle is uniquely defined by the length of his 3 sides) with one common side, and therefore one unique polygon.
As stated the problem can have many solutions.
For example:
Let the quadrilateral $ABCD$ such that $AD=DB=BC=x$ and $AC=\sqrt{3}x$. See figure 1.
Figure 1
The angle $b$ depends on $a$ as the expression:
$$b = \frac{a}{2}- \frac {\pi}{2}+ \arccos(- \frac{1}{2 \sin {\frac{a}{2}}}+ \sin {\frac{a}{2}})$$
So we have many solutions, for example: $a=\frac{\pi}{3}$ and $b=\frac{\pi}{3}$ or $a=\frac{\pi}{4}$ and $b=\frac{\pi}{2}$ as shown in figure 2.
Figure 2
Best Answer
Here's a brute-force derivation ...
$$\begin{align} |\square PQRS| &= |\triangle PQR| + |\triangle PRS| \\[4pt] &=\frac12 p q \sin \angle Q + \frac12 d r \sin(\angle R-\angle PRQ) \\[4pt] &=\frac12 \left(\;p q \sin \angle Q + r \left(\;\sin\angle R\cdot d \cos\angle PRQ - \cos\angle R\cdot d\sin\angle PRQ\;\right) \;\right)\\[4pt] &=\frac12 \left(\;p q \sin \angle Q + r \left(\;\sin\angle R\cdot(q-p\cos\angle Q) - \cos\angle R\cdot p\sin\angle Q\;\right) \;\right)\\[4pt] &=\frac12 \left(\;p q \sin \angle Q +qr\sin\angle R - pr \left(\;\sin\angle R \cos\angle Q + \cos\angle R\sin\angle Q\;\right) \;\right)\\[16pt] &=\frac12 \left(\;p q \sin \angle Q +qr\sin\angle R - pr \sin(\angle Q+\angle R)\;\right)\\ \end{align}$$
In the specific case of $p=r$, the formula can be manipulated thusly: $$\begin{align} |\square PQRS| &= \frac{p}{2}\;\left(\;q(\sin Q + \sin R) - p \sin(Q+R) \;\right) \\[4pt] &= p\sin\frac{Q+R}{2}\;\left(\;q\cos\frac{Q-R}{2}- p \cos\frac{Q+R}{2} \;\right) \end{align}$$