Question 135988
The trouble is, you didn't specify which of the triangle legs you are calling the base.


Assumption:  The longer leg is the base:


{{{drawing(600,400,-1,12,-1,8,
triangle(0,0,0,6,6sqrt(3),0),
locate(3sqrt(3)-.2,-.5,base)
)}}}


A 30-60-90 triangle has sides in the following proportion:  {{{1/2}}}:{{{sqrt(3)/2}}}:{{{1}}}


That means the shorter leg is half the length of the hypotenuse and the longer leg is {{{sqrt(3)/2}}} times the hypotenuse.  And the long leg is then {{{sqrt(3)}}} times the short leg.


Here your long leg is {{{6sqrt(3)}}}, so the short leg, and the height of the triangle must be {{{6sqrt(3)/sqrt(3)=6}}}.


Therefore your area is given by:  {{{A=(6*6sqrt(3))/2=18sqrt(3)}}}


==============================================================


Assumption:  The shorter leg is the base:


{{{drawing(400,600,-1,12,-1,20,
triangle(0,0,0,18,6sqrt(3),0),
locate(3-.2,-.5,base)
)}}}


Here your short leg is {{{6sqrt(3)}}}, so the long leg, and the height of the triangle must be {{{6sqrt(3)*sqrt(3)=6*3=18}}}.


Therefore your area is given by:  {{{A=(18*6sqrt(3))/2=54sqrt(3)}}}


Makes a difference by a factor of 3.