Question 1204108
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This is what the diagram could look like
{{{
drawing(400,400,-4,4,-4,4,
line(-2,-1,2,-1),
line(2,-1,0,2.4641),
line(0,2.4641,-2,-1),
line(0,0.1547,-2,-1),
line(0,0.1547,0,-1),

line(0,-1+0.2,0-0.2,-1+0.2),
line(0-0.2,-1+0.2,0-0.2,-1),

circle(0,0.1547,1.1547),

locate(-2-0.2,-1-0.1,"A"),
locate(2+0.1,-1-0.1,"B"),
locate(0+0.1,2.4641+0.4,"C"),

locate(0+0.1,0.1547+0.4,"D"),
locate(0+0.1,-1-0.1,"E"),

circle(-2,-1,0.05),circle(-2,-1,0.07),circle(-2,-1,0.09),circle(-2,-1,0.11),circle(2,-1,0.05),circle(2,-1,0.07),circle(2,-1,0.09),circle(2,-1,0.11),circle(0,2.4641,0.05),circle(0,2.4641,0.07),circle(0,2.4641,0.09),circle(0,2.4641,0.11),circle(0,0.1547,0.05),circle(0,0.1547,0.07),circle(0,0.1547,0.09),circle(0,0.1547,0.11),circle(0,-1,0.05),circle(0,-1,0.07),circle(0,-1,0.09),circle(0,-1,0.11),

locate(-3.8,-2.8,matrix(1,4,"Diagram","is","to","scale"))

)
}}}

A,B,C = vertices of the equilateral triangle
D = center of the inscribed circle
E = midpoint of AB



Triangle ADE is a right triangle. More specifically, it is a 30-60-90 triangle.
This is because angle CAB = 60 is bisected to help form angle DAE = 30 degrees.
Also, angle ADE = 60 degrees.


The perimeter of the equilateral triangle is 10.8 units.
Each side must be (10.8)/3 = 3.6 units
AB = 3.6
BC = 3.6
AC = 3.6


Because E is the midpoint of AB, we then know
AE = AB/2 = (3.6)/2 = 1.8


Segment AE is the longer leg of the 30-60-90 triangle ADE (notice it's opposite the 60 degree angle).


For any 30-60-90 triangle we have this template
{{{longLeg = shortLeg*sqrt(3)}}}


In this particular case it means
{{{AE = DE*sqrt(3)}}}


Isolating DE gets us 
{{{DE = (AE)/(sqrt(3)) = (1.8)/sqrt(3) = 1.8*(sqrt(3)/3) = 0.6*sqrt(3) = 1.03923}}}
Which is the approximate radius of the inscribed circle.


Round this value however needed. Or you can stick to the exact value {{{0.6*sqrt(3)}}} to avoid worrying about rounding.


Side note: 0.6 = 3/5
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