Question

Consider a square with vertices at (1,1), (-1, 1), (-1, -1) and ( 1, -1). Let be the region consisting of all points inside the square which are nearer to the origin than to any edge. Sketch the region and find its area.

Answer 1

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, * 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 * is the region lying inside the four parabolas

where * is the shaded region.

Now, * is symmetrical in all four quadrants, therefore,

• = 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