Question 1009441
<pre>
We will need: 

1. the height of the triangle so that we can find its area, and 
2. the radius of the circle so that we can find its area.

Then we can subtract to find the other part of the circle. 

{{{drawing(300,300,-1.2,1.2,-1.2,1.2,

line(0,1,-0.8660254,-0.5),
line(-0.8660254,-0.5,0.8660254,-0.5),
line(0.8660254,-0.5,0,1),locate(-.6,.3,12),locate(.47,.3,12),
circle(0,0,1),locate(0,-.5,12) )}}}

First we find the height of the triangle. 
We draw the altitude h.

{{{drawing(300,300,-1.2,1.2,-1.2,1.2,

line(0,1,-0.8660254,-0.5),
line(-0.8660254,-0.5,0.8660254,-0.5),locate(-.6,.3,12),locate(.47,.3,12),
line(0.8660254,-0.5,0,1),locate(-.4,-.5,6),locate(.4,-.5,6),
circle(0,0,1),locate(.02,.25,h),
line(0,1,0,-.5) )}}}

By the Pythagorean theorem,

{{{6^2+h^2=12^2}}}
{{{36+h^2=144}}}
{{{h^2=108}}}
{{{h=sqrt(108)}}}
{{{h=sqrt(36*3)}}}
{{{h=6sqrt(3)}}}

The area of the equilateral triangle

{{{A=expr(1/2)base*height}}}

{{{A=expr(1/2)(12)(6sqrt(3))}}}

{{{A=36sqrt(3)}}}

Next we draw the radii of the circle to the vertices
of the triangle.  Notice that this divides the height of
the triangle into the radius r and the apothem a which is
the height of the lower triangle:

{{{drawing(300,300,-1.2,1.2,-1.2,1.2,

line(0,1,-0.8660254,-0.5),
line(-0.8660254,-0.5,0.8660254,-0.5),locate(-.6,.3,12),locate(.47,.3,12),
line(0.8660254,-0.5,0,1),locate(-.4,-.5,6),locate(.4,-.5,6),
circle(0,0,1),locate(.02,-.2,a),locate(.02,.5,r),locate(-.4,-.1,r),
line(0,1,0,0),
line(-0.8660254,-0.5,0,0),
line(0.8660254,-0.5,0,0),



line(0,1,0,-.5) )}}}

We know that {{{r+a = h}}}, so {{{a=h-r}}}
We know that {{{h=6sqrt(3)}}}
So {{{a=6sqrt(3)-r}}}

By the Pythagorean theorem, 

{{{6^2+a^2=r^2}}}
{{{36+(6sqrt(3)-r)^2=r^2}}}
{{{36+(36*3-12sqrt(3)*r+r^2)=r^2}}}
{{{36+108-12sqrt(3)*r+r^2=r^2}}}
{{{144-12sqrt(3)*r+r^2=r^2}}}

Subtract {{{r^2}}} from both sides:

{{{144-12sqrt(3)*r=0}}}

Solve for r:

{{{144=12sqrt(3)*r}}}

{{{144/(12sqrt(3))=r}}}

{{{12/sqrt(3)=r}}}

The area of the circle is given by

{{{A=pi*r^2}}}

{{{A=pi*(12/sqrt(3))^2}}}

{{{A=pi*(144/3)}}}

{{{A=pi*48}}}

{{{A=48pi}}}

So the area of the other part of the circle besides the
equilateral triangle is found by subtracting the area of 
the equilateral triangle from the area of the circle:

Answer: {{{48pi-36sqrt(3)}}}, or about 88.4

If we like, we can factor out 12

Answer:  {{{12(4pi-3sqrt(3))}}} but that isn't necessary.

Edwin</pre>