Answer
$A(r) = (\frac{\pi}{3}- \frac{\sqrt3}{4})x^2$
Work Step by Step
In equilateral triangle, circumcenter, incenter coincide
So OA is also angle bisector.
This gives $r = \frac{x}{\sqrt3}$
Area of circle but outside the triangle A(r) =
area of circle - area of triangle
= $\pi r^2 - \frac{1}{2}x\frac{\sqrt3}{2}x$
= $\pi (\frac{x}{\sqrt3})^2 - \frac{\sqrt3}{4}x^2$
= $(\frac{\pi}{3}- \frac{\sqrt3}{4})x^2$
$A(r) = (\frac{\pi}{3}- \frac{\sqrt3}{4})x^2$
