Question 1104761
Make an altitude from a vertex to middle of opposite side.  Now the equilateral triangle is two congruent right triangles of the 30-60-90 type.  Hypotenuse is {{{4pi}}} and the short leg is {{{2pi}}}.



Find the size of the altitude.  Call this y.


{{{y^2+(2pi)^2=(4pi)^2}}}
{{{y^2=16pi^2-4pi^2}}}
{{{y^2=12pi^2}}}
{{{y=sqrt(12pi^2)}}}
{{{y=2pi*sqrt(3)}}} or you could say, {{{y=sqrt(3)*2pi}}}, or {{{2*sqrt(3)pi=y}}}.


AREA OF THE EQUILATERAL TRIANGLE
{{{(1/2)*base*altitude}}}

Take any of the {{{4pi}}} sides as base; you already found altitude.

{{{(1/2)*4pi*2*sqrt(3)pi}}}

{{{2pi^2*2*sqrt(3)pi}}}

{{{highlight(4*pi^3*sqrt(3))}}}, or into whichever way that commutative property for multiplication allows that you want.


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