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

Explanation

Let us consider any point T(x,y) inside the square such that its distance from origin is less than or equal to its distance from any of the edges.

`:." "OTltTP,TQ,TR,TS`
Consider `OTltTP`
`rArr" "sqrt(x^(2)+y^(2))lt1-x`
`rArr" "y^(2)le-2(x-(1)/(2))` (1)
So, point T lies on or inside the parabola `y^(2)=-2(x-1//2)`. having vertex at (a/2,0) and which is concave to the left.
Also, parabola passes through the points `(0,pm1)` on y-axis.
Similarly, we have other inequalities as follows :
For `OTltTQ,"we get "x^(2)le2(y+(1)/(2))`.
For `OTltTR,"we get "y^(2)le2(x+(1)/(2))`.
For `OTltTS,"we get "x^(2)le-2(y-(1)/(2))`.
Common region of all the inequalities is plotted as shown in the following figure.