The diagram below shows a perspective drawing of two squares, with coordinates given-on the drawing-for some of the corners of the squares(the line a the top is the horizon)
The diagram below shows the same two squares, now seen from above, with coordinates given for two of the corners.
Determine the coordinates for the remaining corners of the two squares in the second diagram.
Fortunately, both squares are in the same plane. At first, I thought perhaps they were on separate planes.
Let the square at bottom left be denoted $ABCD$, with $A$ at lower left and running counter-clockwise around from there. Let the other square be denoted by $CEFG$, with the letters again running counter-clockwise.
The lower diagram is a top-down, which shows the "real" positions of the squares. The upper diagram is the perspective drawing, with the camera at infinity; we will distinguish those points by using primes ($'$), as in $A', B'$, etc. Our objective is to find, for any point $P(x, y)$ in the "real" drawing, a mapping between it and the corresponding point $P'(x', y')$ in the perspective drawing.
We begin by observing that $\overline{A'D'}$ and $\overline{B'C'}$ must point to the same point at the horizon, since they are in "reality" parallel sides of a square. It should be clear that this point is at $(0, 1)$ in the perspective drawing, hence the horizon in that drawing has the equation $y' = 1$. We can therefore formulate the mapping as follows. We introduce a scaling factor $\alpha$ that depends on a point's "real" $y$-position in some way, and then
$$ x' = \alpha x $$ $$ y' = 1-\alpha $$
where the $1$ comes from the horizon at $y' = 1$. The key point to apply this to is $F(x, y)$. We do not know $x$ and $y$, but we do know that they map, according to the above transformations, to the point $F'(6/11, 6/11)$. That is:
$$ \frac{6}{11} = \alpha x $$ $$ \frac{6}{11} = 1-\alpha $$
so we know that $\alpha = 1-(6/11) = 5/11$, and then
$$ x = \frac{6/11}{5/11} = 6/5 $$
If we assume that the squares are both unit squares (not explicitly given, but it seems likely given what follows), we can then determine $y$ based on the Pythagorean theorem, by observing that the diagonal distance $CF = \sqrt{2}$, with $C$ at $(1, 1)$:
$$ (x-1)^2+(y-1)^2 = (\sqrt{2})^2 $$ $$ \left(\frac{6}{5}-1\right)^2+(y-1)^2 = 2 $$ $$ \frac{1}{25}+(y-1)^2 = 2 $$ $$ (y-1)^2 = \frac{49}{25} $$ $$ y-1 = \frac{7}{5} $$
since $F$ is clearly above $C$, so $y = 12/5$. That is, $F$ is at coordinates $(6/5, 12/5)$. Since we now know two points of the upper square, $C$ and $F$, you should be able to complete the problem by finding the coordinates of $E$ and $G$. First find the center of the upper square by taking the median of the segment $\overline{CF}$, then figure out the proper $x$ and $y$ offsets from that center to the points $E$ and $G$. (It may help to draw this out on a piece of graph paper.)