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 Sketch the region S & find its area

• A. $\frac{1}{3}(8\sqrt{2}+20)sq.units$
• B. $\frac{1}{3}(16\sqrt{2}-20)sq.units$
• C. $\frac{1}{3}(8\sqrt{2}-20)sq.units$
• D. $\frac{1}{3}(16\sqrt{2}+20)sq.units$

Answer 1

The correct option is B $\frac{1}{3}(16\sqrt{2}-20)sq.units$

Distance of point P from origin is less then distance of P from y = 1
$\sqrt{h^2+k^2}<k-1;\sqrt{h^2+k^2}<-k-1$
$\Rightarrow x^2+y^2<{(y-1)}^2;x^2+y^2<y^2+2y+1$
$\Rightarrow x^2<-2(y-\frac{1}{2});x^2<2(y+\frac{1}{2})$
similarly $y^2<-2(x-\frac{1}{2});y^2<2(x+\frac{1}{2})$
$\Rightarrow y=\frac{x^2-1}{-2}ory=x=\frac{x^2-1}{-2}$
$\Rightarrow x^2+2x-1=0\Rightarrow x=-1\pm \sqrt{2}$
$A=8{\int }_0^{\sqrt{2-1}}[\frac{1-x^2}{2}-\sqrt{2}+1]dx+4{(\sqrt{2}-1)}^2=\frac{16\sqrt{2}-20}{3}$