#### Answer

Area = $10A-A^{2}$, $0\lt A\lt10$

#### Work Step by Step

Let the lengths of the sides be $A$ and $B$, where $A\neq B$
Perimeter = $2A+2B=20$
$A+B=10, B=10-A$
Area = $AB=A(10-A)=10A-A^{2}$
$\because$ Lengths are positive
$A\gt0,$
$B=10-A\gt0, A\lt10$
$\therefore 0\lt A\lt10$