How to solve the following ??

Question

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 $S$ .

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Answer 1 · 2017-10-06T14:14:49

See below.

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

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