[tex]\bf A=\cfrac{1}{2}20h\implies A=\cfrac{20h}{2}\implies A=10h[/tex]
now, the range for that equation is all real numbers, is a linear equation and therefore will pass the "vertical line test" and thus be a one-to-one function, and therefore will have an inverse function.
for a function that has an inverse function, the range of the original function is exactly the same as the domain for its inverse. Therefore, if the range of that function is all real numbers, then the domain of its inverse is just that.
what's the inverse, well, let's do the switcharoo.
[tex]\bf A=10h\qquad inverse\implies \boxed{h}=10\boxed{A}\impliedby A^{-1}(h)[/tex]
we are not solving for "h", because, well, is already solved for "h".
