Question 1189025
<br>
{{{drawing(600,600,-1,7,-1,7
,line(0,0,4.9,0),line(0,0,0,4.9),line(4.9,0,0,4.9)
,line(1.79,0,2.45,2.45),line(0,1.79,2.45,2.45),line(1.79,0,0,1.79)
,locate(-.2,4.9,A),locate(0,0,B),locate(4.9,0,C),locate(-.2,1.79,D),locate(1.79,0,E),locate(2.65,2.45,F)
)}}}<br>
Given:
AC=4*sqrt(3)
F is the midpoint of AC, so AF=FC=2*sqrt(3)
DEF is equilateral<br>
To find: The perimeter of DEF<br>
Draw BF intersecting DE at G:<br>
{{{drawing(600,600,-1,7,-1,7
,line(0,0,4.9,0),line(0,0,0,4.9),line(4.9,0,0,4.9)
,line(1.79,0,2.45,2.45),line(0,1.79,2.45,2.45),line(1.79,0,0,1.79)
,red(line(0,0,2.45,2.45))
,locate(-.2,4.9,A),locate(0,0,B),locate(4.9,0,C),locate(-.2,1.79,D),locate(1.79,0,E),locate(2.65,2.45,F)
,red(locate(1,1,G))
)}}}<br>
BF bisects DE; and the length of BF is 2*sqrt(3) -- same as AF and FC.<br>
BF divides DEF into two 30-60-90 right triangles.<br>
Let x be the length of EG; then the side length of DEF is 2x, and FG is x*sqrt(3).<br>
BGE is an isosceles right triangle, so the length of BG is also x.<br>
Now we have BF = BG+GF:<br>
{{{2*sqrt(3)=x+x*sqrt(3)}}}
{{{2*sqrt(3)=x(1+sqrt(3))}}}
{{{x=2*sqrt(3)/(1+sqrt(3))}}}
{{{x=(2*sqrt(3))(1-sqrt(3))/((1+sqrt(3))(1-sqrt(3)))}}}
{{{x=(2*sqrt(3)-6)/(1-3) = (6-2*sqrt(3))/2 = 3-sqrt(3)}}}<br>
Finally, the perimeter of DEF is {{{6x=6(3-sqrt(3))}}}<br>
ANSWER: A {{{6(3-sqrt(3))}}}<br>