A circle with radius r, is inside an equilateral triangle, with sides x cm

First find the radius of the circle   area = pi r^2   r = 5.64 cm

Then   tan 30 = 1/2 / (sqrt(3) /2)) = r/(1/2x)   where x is side length

1/2 / sqrt 3/ 2 =  r/(1/2 x)

1/2 x  * .57735 = r   = 5.64    x = 19.53 ~~~ 20 cm (two sig-figs  

 See diagram below:

Ooops ...forgot diagram (and I found an error...fixed)

The radius of the circle  is :     sqrt [ (100) / pi ]  =  10 / sqrt (pi)

We can find 1/2 of the length of the side of the triangle thusly :

[ (1/2)x ] /  sin (60)  =  [10/ sqrt (pi)]  / sin (30)

x / [ 2 sqrt (3) /2 ]  =  2 * 10 / sqrt(pi)

 x / sqrt (3)   =  20/ sqrt(pi)

x =  20 sqrt (3)  / sqrt (pi)   =   20  sqrt (3/pi) ≈ 19.54 cm

( 20 cm  rounded to 2 sig figs )

Here is the diagram I sketched:

The radius of the circle  is :     sqrt [ (100) / pi ]  =  10 / sqrt (pi)

We can find 1/2 of the length of the side of the triangle thusly :

[ (1/2)x ] /  sin (60)  =  [10/ sqrt (pi)]  / sin (30)

x / [ 2 sqrt (3) /2 ]  =  2 * 10 / sqrt(pi)

 x / sqrt (3)   =  20/ sqrt(pi)

x =  20 sqrt (3)  / sqrt (pi)   =   20  sqrt (3/pi) ≈ 19.54 cm