Orthographic projection of a rectangle 
Rectangle $ABCD$ is projected using orthographic projection onto a plane that makes a known angle with the plane of the rectangle, as shown in the figure above.  Can the lengths of the sides of the rectangle be determined using the projected image?
(Note: This is not specific to the parallelogram given in the figure.)
 A: No.
This can be shown intuitively with a simple example:
Suppose that the angle is $\frac{\pi}{3}=60°$ (such that the cosine of this angle is $\frac{1}{2}$) and that the projected figure is a rectangle with sides $1$ and $2$.
From the fact that the projected figure is a rectangle, we can conclude that the intersection line of the two planes must be parallel to one of the sides of the rectangle. But we don't know which.

*

*If it is parallel to the long side, then the original rectangle must be a square with side length $2$.

*If the intersection line is parallel to the short side, then we know that the original rectangle has side lengths of $1$ and $4$.

But we have no way to distinguish between these two cases.
A: A rough 3d Geogebra sketch of related projection geometry of a rectangular prism.

A rectangular prism between planes
$$ (\pm 3,\pm4,0), (\pm 3,\pm4,5)$$
when cut by a plane of arbitrary inclination produces a section with  parallelogram boundary.
A: They cannot be determined.
Proof: Assume that the angle is known to be $\theta.$ Take a square with one of its sides parallel to the plane of projection and with side length $\sec(\theta).$ It would project to a $1 \times \sec(\theta)$ rectangle.  A rectangle with the short side parallel to the plane of projection that has dimensions $1 \times \sec^2(\theta)$ also projects to a rectangle which has dimensions $1 \times \sec(\theta)$. Since these two different rectangles project to the same rectangle, it is impossible to tell which one is the square, so they cannot be determined, even if we can choose the angle $\theta.$
A: No.
An orthographic projection from one plane to another plane can be imagined as squashing one direction by a factor of $\cos\theta.$ Therefore, we can undo an orthographic projection by stretching in one direction by a factor of $\sec\theta.$
Here's a Desmos graph that shows my counterexample. The points are $(-0.5,0.5s), (0.5,-0.5s), (-\frac{2-k}{2\sqrt7},\frac{2+k}{2\sqrt7}s),$ and $ (\frac{2-k}{2\sqrt7},-\frac{2+k}{2\sqrt7}s).$ Varying $k$ stretches the four points by a factor of $k$ in the direction of the line $x = y$, and preserving the line $x = -y.$ Varying $s$ stretches along the $y-$axis and preserves the $x-$axis. As can be seen in the graph, stretching along either of those directions by a factor of $\sqrt3$ by changing $k$ or $s$ to $\sqrt3$ makes the given parallelogram into a rectangle. Since we can undo the orthographic projection in two different ways to get a rectangle, the rectangle is not uniquely determined.
(Footnote: The angle $\theta$ in my counterexample is $\arccos \sqrt\frac{1}{3}.$)
