#### Answer

18 square units

#### Work Step by Step

Solving the system
$\left\{\begin{array}{l}
x^{2}+y^{2}=41\\
xy=20
\end{array}\right.$
graphically (see below)
we find the solution set to be the points
(4,5), (5,4), (-5.-4), and (-4.-5).
The distance from (4,5) to (5,4) is
$a=\sqrt{(5-4)^{2}+(4-5)^{2}}=\sqrt{2}$
The distance from (4,5) to ($-5,\ -4$) is
$b=\sqrt{(-5-4)^{2}+(-4-5)^{2}}=\sqrt{2\cdot 81}=9\sqrt{2}$
a and b are the sides of the rectangle described in the problem.
Its area is
$ab=\sqrt{2}\cdot 9\sqrt{2}=9\cdot 2=18$ square units
.