Question 855606
The center of the circle inscribed in the triangle is the incenter, the point at the same distance from all 3 sides.
It can be found as the intersection of the angle bisectors (all 3 intersect there, but we can find the incenter by drawing any two of them).
 
If I were the artist my triangle would be an equilateral triangle.
{{{drawing(300,300,-6,6,-2,10,
green(rectangle(0,0,0.4,0.4)),triangle(-5,0,5,0,0,8.66),
green(circle(0,2.89,0.1)),green(line(0,2.89,5,0)),
green(line(0,2.89,-5,0)),green(line(0,2.89,0,8.66)),
red(circle(0,2.89,2.89)),
green(arrow(0,0,0,15))
)}}}
An equilateral triangle with a perimeter of 30 inches would have a side length of 10 inches.
In an equilateral triangle, the height is {{{sqrt(3)/2}}} times the length of the side.
In an equilateral triangle, connecting the incenter to the vertices splits the triangle into 3 congruent triangles.
They all have the same base as the equilateral triangle,
and {{{1/3}}} of the area of the equilateral triangle.
As a consequence their heights are {{{1/3}}} of the height of the equilateral triangle.
Those heights, measured perpendicular to the sides to the incenter are the radius of the inscribed circle.
In my triangle the side measures {{{10}}}inches;
the height of the equilateral triangle is {{{10sqrt(3)/2=5sqrt(3)}}} inches,
and the radius of the inscribed circle is {{{(1/3)*(5sqrt(3))=5sqrt(3)/3}}} .
The circumference of the circle would be
{{{2*pi*5sqrt(3)/3=10pi*sqrt(3)/3=18.14}}} inches (rounding).
If the artist uses 30 inches of silver wire for the triangle,
and 18.14 inches of copper wire for the circle,
she used {{{30-18.14=highlight(11.86)}}} more inches of silver wire than of copper wire.