Question 1208457
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Hint:
{{{
drawing(400,400,-5,5,-5,5,
circle(-2,0,0.05),circle(-2,0,0.07),circle(-2,0,0.09),circle(-2,0,0.11),circle(2,0,0.05),circle(2,0,0.07),circle(2,0,0.09),circle(2,0,0.11),circle(0,3.4641016151,0.05),circle(0,3.4641016151,0.07),circle(0,3.4641016151,0.09),circle(0,3.4641016151,0.11),circle(0,-4,0.05),circle(0,-4,0.07),circle(0,-4,0.09),circle(0,-4,0.11),circle(0,0,0.05),circle(0,0,0.07),circle(0,0,0.09),circle(0,0,0.11),
line(-2,0,2,0),line(2,0,0,3.4641016151),line(0,3.4641016151,-2,0),line(0,3.4641016151,0,-4),line(0,-4,-2,0),line(0,-4,2,0),
locate(-2-0.2,0-0.1,"A"),locate(2+0.2,0,"B"),locate(0,3.4641016151+0.6,"C"),locate(0,-4-0.2,"D"),locate(0+0.2,0-0.2,"E"),locate(-1.16,-0.26+0.8,"x"),locate(1.16,-0.26+0.8,"x"),locate(0.16,1.54,x*sqrt(3)),locate(-1.38,1.92+0.2,2x),locate(1.38,1.92+0.2,2x),locate(0.14,-1.62+1,18-x*sqrt(3)),locate(-1.34-0.5,-1.96,"10"),locate(1.52,-1.66,"10"),

line(-0.34,0,-0.34,-0.34),line(-0.34,-0.34,0,-0.34),

locate(-4.5,-4,matrix(1,2,"Diagram","not")),
locate(-4.5,-4.5,matrix(1,2,"to","scale"))
)
}}}
ABC is an equilateral triangle with side length 2x. 
D is the point outside of the equilateral triangle such that DA = 10, DB = 10, and DC = 18.
E is the midpoint of segment AB.
AE = x and EB = x
The height of the equilateral triangle is EC = x*sqrt(3)
DE = DC-EC = 18-x*sqrt(3)
AED is a right triangle.
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