Question 259341
Since it's a right triangle, it observes the Pythagorean theorem.
{{{A^2+(H/2)^2=H^2}}}
{{{A^2=H^2-(H^2/4)}}}
{{{A^2=(3/4)H^2}}}
{{{A=(sqrt(3)/2)*H}}}
This only works if 
{{{H=Z/sqrt(3)}}} to cancel out the sqrt(3) in the numerator and Z is a whole number greater than or equal to 2. 
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So yes, 
{{{H=2/sqrt(3)}}}
would then give you a short side of 
{{{A=sqrt(3)/2*(2/sqrt(3))=1}}}