Question 1180153
The area of a circle inscribed in an equilateral triangle with side b is:

{{{ A[circle] = pi*b^2/12 }}}<br>

Plug in b=10cm:  {{{ A = 26.18cm^2 }}}

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The long way (I'm sure there are other ways to solve this, this is the first method that came to mind):  

Draw the triangle with inscribed circle.  

Label the radius r, and draw an altitude of the triangle.  This altitude has length 2r + d (where d is the portion of the altitude outside of the circle) and using Pythagorean Theorem:
{{{ 2r + d = sqrt(10^2-5^2) = 5*sqrt(3) }}}

Now draw another radius from the center of the circle to the midpoint of one of the sides of the triangle (it will meet right where the circle is tangent to one side of the triangle).  This forms a small right triangle with sides r, r+d, and 5.    Using Pythagorean Theorem on this triangle:
  {{{r^2+5^2 = (r+d)^2}}}
which reduces to
  {{{ d(d+2r)-25 = 0 }}}
  
Substitute {{{ d+2r = 5sqrt(3) }}} from earlier into this last equation and
solve for d:  {{{ d = 5sqrt(3)/3 }}} and therefore you can solve {{{ 2r = 10*sqrt(3)/3 }}} or  {{{ r = 5sqrt(3)/3 }}}

{{{ A[circle] = pi*(5sqrt(3)/3)^2 = 26.18cm^2 }}}

It is interesting to note  d = r, so the altitude of the triangle is 3r.