Heron kills it

proof without words

Using Heron's formulae for triangle area with sides $x,y,z$ where the area is $\frac{1}{4}\sqrt{2x^2y^2+2y^2z^2+2z^2x^2-x^4-y^4-z^4}$ , we subsitute $x=\sqrt{a^2+16b^2},y=\sqrt{9a^2+4b^2},z=\sqrt{4a^2+4b^2}$ , expand and then reduce our expression to get an area of:

$\frac{1}{4}\sqrt{2((a^2+16b^2)(9a^2+4b^2)+(9a^2+4b^2)(4a^2+4b^2)+(4a^2+4b^2)(a^2+16b^2))-((a^2+16b^2)^2+(9a^2+4b^2)^2+(4a^2+4b^2)^2)}$

$=\frac{1}{4}\sqrt{400a^2b^2}=5ab=30$

We thus desire to find all ordered pairs of positive integers $(a,b)$ where $ab=6$ . There are then $\boxed{4}$ such pairs: $(1,6),(2,3),(3,2),(6,1)$ .