For convenience of explanation, let the two vertices below the square be A and B, the irregular figure be Y, the uppermost point of Y be C, and the leftmost, bottommost and rightmost points be D, E, F, E and F respectively. The area of ADE connection is recorded as x, and when it is connected with ABC, it can get △ABC. Because AB=BC=AC=a, this is a regular triangle, and then we can draw that every small arc in the diagram is equal, which is easy to verify.

It is easy to find that the area of this triangle is the root number 3*a? /4, so we can find that the area enclosed by arc AC and line segment AC is a? π/6-3*a？ /4, and because it is said that every small arc in the figure is equal, the area of ACF connection is a? π/ 12, so 2x+y=a? π/2-a？ =2*(a？ π/6-3*a？ /4)+a？ π/ 12+x, the above two equations can get the value of y, y=( 1+π/3- radical number 3)a? . The first two buildings are idiots. Don't listen to them. This must be right!