jmc

Let the side lengths of the rectangle be $a$ and $b$ with $a\le b$. Then $ab=10(a+b).$ Expanding and moving all the terms to the left hand side gives $ab-10a-10b=0.$ We apply Simon's Favorite Factoring Trick and add $100$ to both sides to allow us to factor the left hand side: $$ab-10a-10b+100=(a-10)(b-10)=100$$From this, we know that $(a-10,b-10)$ must be a pair of factors of $100$. Consequently, the pairs $(a,b)$ that provide different areas are $(11,110),$ $(12,60),$ $(14,35),$ $(15,30),$ and $(20,20)$. There are therefore $\boxed{5}$ distinct rectangles with the desired property.