[bar94]

Not sure if everything is correct, seems a bit too easy -

Let y be the horizontal width of the bottom-left-most rectangle, and x be the horizontal width of each of the rectangles directly adjacent on the right to the first one, so that the rectangle marked at the top right as having a vertical width of 3 now has a horizontal width of x + y.

Then, the area of each rectangle is 3x + 3y, as all rectangles have the same area.

It follows that the vertical width of the bottom-left-most rectangle is (3x + 3y)/y, and that the vertical width of each of the two stacked rectangles is (3x + 3y)/x.

Note now that the stacked rectangles have the same horizontal width, and therefore must have the same vertical width. Thus, the vertical width of the bottom-left-most rectangle must be 2 times the vertical width of each of these stacked rectangles, so (3x + 3y)/y = 2*(3x + 3y)/x.

From this, we get 1/y = 2/x (noting that 3x + 3y cannot be be zero), so x = 2y.

This means that the vertical width of the bottom-left-most rectangle is then (3(2y) + 3y)/y = (6y + 3y)/y = 9y/y = 9.

Therefore, the entire left side width is 3 + 9, or 12 units. Since the overall shape is a square, the entire square's area must be 144 square units.

[kwstas]

I'm not really good at parsing information over text so I had to lay your solution out over the problem to get it.

https://i.imgur.com/OekI7jZ.png

> Note now that the stacked rectangles have the same horizontal width, and therefore must have the same vertical width.

This is the most important thing to use and what, were it included in the system in the OP, would simplify it quite a bit with not much more complexity. But the value of the approach I guess is the use of only pre-defined constraints.

Another route is to use that h_1 * (x+y) = 3A = 9 * (x+y) so h_1 = 9

[Tarq0n]

In the image the green square is actually less tall than the pink, which is really misleading.