Geometry Problem: Find the Area of the Intersection Between Two Squares 

Draw perpendiculars from the center of the smaller square to the top and right sides of the smaller square. This gives us three polygons inside the smaller square: two right triangles, one shaded and one not, and a shaded quadrilateral between them. Given the equal length of the two segments we drew and the 90° angles at the top and right sides of the smaller square and the corner of the larger square, it is simple to show that the shaded triangle and the unshaded triangle are congruent, and thus have the same area. Our red shaded area consists of the quadrilateral and the shaded triangle, but the unshaded triangle and the quadrilateral combine to form one quarter of the smaller square. As the triangles are congruent, these areas are the same. Aᵣ = s²/4 = 2²/4 = 4/4 = 1 sq unit And the side of the side length 3 square isn't entirely unnecessary. This only works if the rotating square has sides of at least a specific size, that size being √2 times half the length of the static square's sides. That's s/√2, or in the case of this problem, √2 in length. If the sides of the rotating square aren't at least this long, then all of the area of the static square between the rotating square's adjacent sides won't necessarily be overlapped. Note thst the "larger" square doesn't actually have to be larger.

My intuition is that it is always exactly 1 because they are obviously 1 when the corresponding sides are parallel and 1 when they are at 45°.

DUH (2x2)/4=1 but your thumb tag was very misleading.

You’re solution only holds true if the corner of large square lies in center of small square . The problem never stated that fact . U assumed the corner point was centered in the small square