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:





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:

  

which reduces to

  

  

Substitute  from earlier into this last equation and

solve for d:   and therefore you can solve  or  







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







 

Answer by ikleyn(52832)   (Show Source): You can put this solution on YOUR website!
 .


The altitude of this triangle (any one of the three its altitudes) is  {{10*(sqrt(3)/2)}}} cm long.



The center of the inscribed circle in this case is the common intersection point of three altitudes and 

the radius of the inscribed circle is    of the altitude length.


So, the radius of the inscribed circle is  r =  = .


Then the area of this circle is   =  =  =   square units.






 

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