How to solve the following ??

Consider a square with vertices at (1,1);(−1,1);(−1,−1);(1,−1). Let S be the region consisting of all points inside the square which are nearer to the origin than to any edge. Find the area of region S.

1 Answer
Oct 6, 2017

See below.

Explanation:

The sought region is the interior of the boundaries given by

sqrt(x^2+y^2) le x+1
sqrt(x^2+y^2) le 1-x
sqrt(x^2+y^2) le y+1
sqrt(x^2+y^2) le 1-y

The area computation is left as an exercise.

Attached a region plot

enter image source here

NOTE:

The intersection region boundary point at the first quadrant is

sqrt2-1,sqrt2-1 and the area is given by

S= (2 (sqrt[2] - 1))^2 +4(int_(1-sqrt2)^(sqrt2-1)(1-x^2)/2 dx - 2(sqrt2-1)^2) = 4/3(4sqrt2-5)