Perfect Perimeters

We can "extend" the sides to fit the entire square. Notice that L, S, T, and + all fit exactly, but H and U have extra parts (marked red in the picture). Thus the answer is $4$ .

Perimeter means sum of all the sides. All we need to see is which sides of the symbols equals what proportion of a side of the square paper. We see that only four symbols L,S,T,+ contributes to every portion of the sides of the square paper. And hence their perimeter is same as the square paper. The other two symbols have greater perimeter than the square paper. So the answer is 4!

Only when all the blank rectangles in the corners.

The lengthy foolproof method. Give the square a metric value, say 5cm, therefore the perimeter = (5 + 5) * 2 = 20. Calculate the L shape perimeter as 5 + 1 + 4 + 4 + 1 + 5 = 20. Calculate the H shape perimeter as 5 + 1 + 2 + 3 + 2 + 1 + 5 + 1 + 2 + 3 + 2 + 1 = 28. Complete for all shapes and produce the answer of 4.

Whenever you can draw a vertical or horizontal line that cuts the external boundary of the shape more than twice, you know that this shape has not the same perimeter as the square. By the way, this only works if all sides of the big square are touched by the internal shape. I think proof is pretty obvious and I'm writing on a smartphone. But if you really want one, I can write one in the comments, let me know! (I'm inconfortable affirming something without proof ^^) I strongly believe that the reciprocal is also true and we have an equivalence between both assertions but I have no good proof now.

Assume that all the angles of the colored shape are right angles. The perimeter segments that are shared by the colored shape and the square are, by definition, equal. If all of the adjacent nonshared segments of a colored shape have 2 segments, then they form rectangle(s) with the corresponding 2 segments of the perimeter of the square, and thus have the same total perimeter, and this is true for 4 of the shapes: L, S, T, and +. If any of the adjacent nonshared segments of a colored shape have 3 segments, then they form 1 or more rectangles with the corresponding 1 segment of the perimeter of the square, and thus have a greater total perimeter, and this is true for 2 of the shapes: H and U. So, the answer to the problem is 4.

If you don't assume that all the angles of the colored shapes are right angles, then I don't believe you can solve the problem, because the interior 2 sides of the 4 sided polygons could be different lengths than the exterior sides.

You can solve it just by expanding the lines to the square limit, after doing that, you obtain 4 figures as the answer...

I added each perimeter of every shape and only 4 matches the perimeter of the square