There is a quadrilateral with equal-length for opposite sides but the diagonals are different (and I hope the word parallelogram is correct here), what would be the biggest rectangle I can inscribe, and why?
To put this in practical terms:
Someone cut a board and checked that the opposite sides have equal length. However when checking the diagonals, there are deviations. How to get back to a rectangular shape with minimum area to be cut off?
Note: reassembly is not intended here. If it were, there is a trivial solution that I am not interested in: https://en.wikipedia.org/wiki/Parallelogram#/media/File:Parallelogram_area_animated.gif
Best Answer
I'll try to expand the comment written under the question:
Consider the figure below:
We call $P$ the parallelogram of base $B$ and height $H$ fixed, and $R$ will be the rectangle inscribed in $P$. Of course $\gamma$ is fixed.
Let's evaluate the area of the triangle $1$
Thanks to a well known property of triangles we have $\frac{l}{\sin\gamma} = \frac{x}{\sin( 180 - \gamma - \alpha)}$ then $$h = l \sin \alpha = \frac{x \sin\alpha\sin\gamma}{\sin(180 - \gamma - \alpha)}$$ and the area of that triangle will be $$\text{Area}_1 = \frac{x^2 \sin\alpha\sin\gamma}{2\sin(180 - \gamma - \alpha)}$$
For the trinagle $2$ we have that his height is $H - h$ and the base is $B - x$. In conclusion: $$\text{Area}_2 = \frac{1}{2}(B-x)\Big(H - \frac{x \sin\alpha\sin\gamma}{\sin(180 - \gamma - \alpha)} \Big)$$
In conclusion, to maximise $R$ area we have that $\text{Area}_1 + \text{Area}_2$ must reach its minimum depending on $x$ and $\alpha$.