Question 1209435
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Equilateral triangle ABC has side length s = 1 which leads to area {{{(1/4)s^2*sqrt(3) = (1/4)*1^2*sqrt(3) = sqrt(3)/4}}}


Equilateral triangle ABC
{{{
drawing(400,400,-0.8,0.8,-0.8,0.8,
line(0.0024,0.4754,-0.4975,-0.3906),line(-0.4975,-0.3906,0.5025,-0.3906),line(0.5025,-0.3906,0.0024,0.4754),

locate(0.0024,0.5374, "A"),locate(-0.4975,-0.3906, "B"),locate(0.5025,-0.3906, "C")
)
}}}


Draw the altitude AD which goes from point A to side BC.
The altitude is perpendicular to BC. 
For any equilateral triangle, an altitude is also a median.
D is the midpoint of BC.


{{{
drawing(400,400,-0.8,0.8,-0.8,0.8,
line(0.0024,0.4754,-0.4975,-0.3906),line(-0.4975,-0.3906,0.5025,-0.3906),line(0.5025,-0.3906,0.0024,0.4754),
line(0.0024,0.4754,0.0024,-0.3906),

locate(0.0024,0.5374, "A"),locate(-0.4975,-0.3906, "B"),locate(0.5025,-0.3906, "C"),locate(0,-0.3906, "D")
)
}}}


The last point to introduce is E. 
This is the midpoint of segment BD
{{{
drawing(400,400,-0.8,0.8,-0.8,0.8,
line(0.0024,0.4754,-0.4975,-0.3906),line(-0.4975,-0.3906,0.5025,-0.3906),line(0.5025,-0.3906,0.0024,0.4754),
line(0.0024,0.4754,0.0024,-0.3906),
line(0.0024,0.4754,-0.4975/2,-0.3906),

locate(0.0024,0.5374, "A"),locate(-0.4975,-0.3906, "B"),locate(0.5025,-0.3906, "C"),locate(0,-0.3906, "D"),locate(-0.4975/2,-0.3906, "E")
)
}}}


Because segment EB is half of BD, and BD is half of BC, this makes EB have a length that is (1/2)*(1/2) = 1/4 of BC's length
EC is 3/4 the length of BC


Since triangles ABC and AEC have the same height we can say,
area of triangle AEC = (3/4)*(area of triangle ABC)


area of triangle AEC = {{{(3/4)*(sqrt(3)/4)}}}


area of triangle AEC = {{{(3*sqrt(3))/16}}} square cm exactly


area of triangle AEC = 0.324759526419 square cm approximately
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