Question 17499
For an equilateral triangle,
Area is given by
{{{ A= (s^2)sqrt(3)/4 }}}
where s is the length of the side
<P>
{{{ 30 = s^2*sqrt(3)/4 }}}
{{{ 30*4/sqrt(3) = s^2 }}}
{{{ 3*40/sqrt(3) = s^2 }}}
{{{ sqrt(3)*40 = s^2 }}}
{{{ s= sqrt(40*sqrt(3)) }}}
{{{ s= sqrt(sqrt(40*40*3)) }}}
{{{ s= sqrt(sqrt(4800)) }}}
Clear till here? Good.
Now we have to take factors twice.
So we get 
{{{ sqrt(4800)=40*sqrt(3) }}} ........[first time]
{{{ sqrt(40*sqrt(3))=2*sqrt(5)*sqrt(2)*sqrt(3)}}}......[second time]
{{{ sqrt(40*sqrt(3))=2*sqrt(10*sqrt(3))}}}......[simplified]
Therefore we get our value of 's' as {{{2*sqrt(10*sqrt(3))}}}
<P>
{{{ A= (s^2)sqrt(3)/4 = (1/2)bh }}}
Now base and side are equal,therefore
{{{ (s^2)sqrt(3)/4 = (1/2)bh }}}
{{{ (s^2)sqrt(3)/4 = (1/2)sh }}}
{{{ s*sqrt(3)/4 = (1/2)h }}}
{{{ s*sqrt(3)/2 = h }}}
{{{ h = (sqrt(3)/2)*2*sqrt(10*sqrt(3))}}}
{{{ h = sqrt(3)*sqrt(10*sqrt(3)) }}}
{{{ h= sqrt(3*10*sqrt(3)) }}}
{{{ h= sqrt(30*sqrt(3)) }}}
<P>
So we have our values as
{{{ s = 2*sqrt(10*sqrt(3)) }}}
{{{ h = sqrt(30*sqrt(3)) }}}
<P>
We can verify this using the [A=1/2bh] formula,
Area=1/2*s*h
{{{ (1/2)*2*sqrt(10*sqrt(3))*sqrt(30*sqrt(3)) }}}
{{{ sqrt(10*sqrt(3)*30*sqrt(3)) }}}
{{{ sqrt(300*sqrt(3)*sqrt(3)) }}}
{{{ sqrt(300*3) }}}
{{{ sqrt(900) }}}
{{{ 30 }}}
<P>
And thus we know that our values for s and h are correct.
Hope this helps,
Prabhat