Question 1105588
{{{drawing(300,275,-0.6,0.6,-0.1,1,
triangle(-0.5,0,0.5,0,0,0.866),
triangle(0,0,0.25,0,0,0.866),
rectangle(0,0,-0.05,0.05),
locate(0.5,0,B),locate(-0.53,0,C),
locate(0.24,0,E),locate(-0.01,0,D),locate(-0.01,0.92,A)
)}}}

An equilateral triangle with side length {{{x}}} has a height {{{(sqrt(3)/2)x}}} 
,
and an area {{{(sqrt(3)/4)x^2}}},
so if {{{x}}} represents the length of side AB in cm, {{{(sqrt(3)/4)x^2}}} is the area of ABC in {{{cm^2}}} .
 
The altitude of an equilateral triangle is also a median,
so it splits the triangle into two triangles with the same area.
Those triangles are ABD and ACD, each with an area equal to {1/2}}} of the area of ABC.
Triangle ABD is similarly split into triangles ABE and AED,
each with area equal to {{{1/2}}}of the area of ABD,
meaning {{{1/4}}} of the area of ABC.
Triangle ARC is the union of triangles ABD and AED,
so its area is {{{1/2+1/4=3/4}}} of the area of ABC.
That is
{{{(3/4)(sqrt(3)/4)x^2=(3sqrt(3)/16)x^2}}} .
 
{{{(3sqrt(3)/16)x^2=27sqrt(3)}}}
{{{(3/16)x^2=27}}}
{{{x^2=27*16/3}}}
{{{x^2=9*16}}}
{{{x=sqrt(9*16)=sqrt(9)sqrt(16)r3*4=highlight(12)}}} .
The length of side AB is {{{highlight(12cm)}}} .