The equations of the sides of the square are as follows
AB : y = 1, BC : x = -1,CD : y = -1, DA : x = 1
Let the region be, S and (x, y) is any point inside it.
Then, according to given conditions,
Now, in y2 =1 - 2x and y2 = 1 + 2x, the first equation represents a parabola with vertex at (1/2,0) and second equation represents a parabola with vertex ( -1/2,0).
And in x2 =1- 2y and x2 = 1 + 2y, the first equation represents a parabola with vertex at (0, 1/2) and second equation represents a parabola with vertex at (0, -1/2) .
Therefore, the region S is the region lying inside the four parabolas
where S is the shaded region.
Now, S is symmetrical in all four quadrants, therefore,
S = 4 x area lying in the first quadrant.
Now, y2 = 1- 2x and x2 = 1 - 2y intersect on the line y = x. The point of intersection is E(√ 2 - 1, √ 2 - 1)
Area of the region OEFO = area of Δ OEH + area of HEFH