Mathematics - Straight Lines Question with Solution | TestHub

For interior $(x,y)$ the distances to $OA,OB,AB$ are $y,(x-\sqrt{3}y)/2$ and $(2-x-\sqrt{3}y)/2$. Thus $R$ has $0\le y\le 1/(2+\sqrt{3})$ and $(2+\sqrt{3})y\le x\le 2-(2+\sqrt{3})y$. It is a triangle of base 2 and height $1/(2+\sqrt{3})=2-\sqrt{3}$, hence area $2-\sqrt{3}$.