## College Algebra (6th Edition)

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 .