Question 607062
As with any triangle, {{{ A = (1/2)*b*h }}}
Let the sides of the equilateral triangle = {{{s}}}
Now the formula becomes {{{ A = (1/2)*s*h }}}
Dropping a vertical to the base forms 2 back to back
30-60-90 triangles.
Opposite the 30 is {{{ s/2 }}}
Opposite the 60 is {{{ (s/2)*sqrt(3) }}}
Opposite the 90 is {{{ s }}}
The vertical, {{{h}}}, is opposite the 60 degree angle, so
{{{ A = (1/2)*s*(s/2)*sqrt(3) }}}
{{{ A = (s^2/4)*sqrt(3) }}}
{{{ A = (s/2)^2*sqrt(3) }}}