Question 318866
prob best to draw the triagle to visualize an equalateral triangle meaning all sides the same the radius is 3 and also it disects the base side at a 90 degree angle and the base is seperated in 2 equal segments. so if x= the hypotenuse of the right triagle which is the same as the length of one side of the equalateral triangle then the base of the right triangle has to be {{{(1/2)x}}} now i would do the therom {{{a^2+b^2=c^2}}} where a=6 b=(1/2)x and c=x
the hieght of the right triangle is 2*(radius) so 2*3=6
<p>
{{{a^2+b^2=c^2}}}
{{{6^2+(x/2)^2=x^2}}}
{{{36+(x^2/4)=x^2}}}
{{{144+x^2=4x^2}}}
{{{144=3x^2}}} divide by 3
{{{48=x^2}}} sqrt each side
{{{sqrt(48)=x}}}
{{{sqrt(3*4*4)=x}}}
{{{4(sqrt(3))=x}}}
<p>
so {{{(1/2)bh=a}}} b=base h=hieght a=area
{{{b=4(sqrt(3))}}}
{{{h=6}}}
<p>
{{{(1/2)(4(sqrt(3)))(6)=a}}}
{{{(4(sqrt(3)))(3)=a}}}
{{{highlight(12(sqrt(3))=a)}}}