Question 919088
{{{drawing(200,4400/27,-2.7,2.7,-.7,3.7,
locate(-1.9,0,B),locate(1.75,0,C),locate(-.1,3.4,A),
line(-sqrt(3),0,sqrt(3),0), line(0,3,-sqrt(3),0), line(0,3,sqrt(3),0),
circle(0,1,1) )}}}
<pre>
Draw the altitude AD, which divides the base BC into two equal parts,
BD = DC.

{{{drawing(200,4400/27,-2.7,2.7,-.7,3.7,
locate(-1.9,0,B),locate(1.75,0,C),locate(-.1,3.4,A),
locate(0,0,D),
line(-sqrt(3),0,sqrt(3),0), line(0,3,-sqrt(3),0), line(0,3,sqrt(3),0),
circle(0,1,1),green(line(0,0,0,3)) )}}}

Since TABC is equilateral, AB = BC = 2·BD. 

By the Pythagorean theorem on right triangle BAD,

BDČ+ADČ=ABČ
BDČ+ADČ=(2·BD)Č
BDČ+ADČ=4·BDČ
    ADČ=4·BDČ-BDČ
    ADČ=3·BDČ
     AD=&#8730;<span style="text-decoration: overline">3</span>·BD



The Area of TABC = {{{1/2}}}(BC)(AD)
                 = {{{1/2}}}(2·BD)(&#8730;<span style="text-decoration: overline">3</span>·BD)
                 = 4&#8730;<span style="text-decoration: overline">3</span>·BDČ

We are told that the area of TABC is 4·&#8730;<span style="text-decoration: overline">3</span>
So we have the equation 

      4&#8730;<span style="text-decoration: overline">3</span> = &#8730;<span style="text-decoration: overline">3</span>·BDČ

Divide both sides by &#8730;<span style="text-decoration: overline">3</span>

        4 = BDČ
        2 = BD

Draw AO, where O is the center of the circle.

{{{drawing(200,4400/27,-2.7,2.7,-.7,3.7,
locate(-1.9,0,B),locate(1.75,0,C),locate(-.1,3.4,A),
locate(0,0,D),red(line(-sqrt(3),0,0,1)),locate(.04,1.3,O),
line(-sqrt(3),0,sqrt(3),0), line(0,3,-sqrt(3),0), line(0,3,sqrt(3),0),
circle(0,1,1),green(line(0,0,0,3)) )}}}

OB bisects < ABC which is 60°, so < OBD is 30° and  BOD = 60°,
so TBOD is a 30°-60°-90° triangle and so OB = 2·OD

By the Pythagorean theorem applied to TBOD,

BDČ+ODČ=OBČ
 2Č+ODČ=(2·OD)Č
  4+ODČ=4·ODČ
  4+ODČ=4·ODČ
      4=3·ODČ
      {{{4/3}}}=ODČ
      
The area of a circle is given by

    Area = <font face="symbol">p</font>·(radius)Č
         = <font face="symbol">p</font>·ODČ               
         = <font face="symbol">p</font>·{{{4/3}}}
         = {{{4pi/3}}}

Edwin</pre>