Answer
The area of the rectangle is $18$ square units.
Work Step by Step
The system of nonlinear equations formed is:
$\left\{ \begin{align}
& {{x}^{2}}+{{y}^{2}}=41 \\
& xy=20
\end{align} \right.$
We use the substitution method to solve the system of equations.
Now, consider Equation (II):
$\begin{align}
& xy=20 \\
& x=\frac{20}{y}
\end{align}$
Substitute the value of x in equation (I) to find the value of y:
$\begin{align}
& {{x}^{2}}+{{y}^{2}}=41 \\
& {{\left( \frac{20}{y} \right)}^{2}}+{{y}^{2}}=41 \\
& \frac{400}{{{y}^{2}}}+{{y}^{2}}=41 \\
& {{y}^{4}}-41{{y}^{2}}+400=0
\end{align}$
Factorize the above equation to solve for the solutions as shown below:
$\begin{align}
& {{y}^{4}}-41{{y}^{2}}+400=0 \\
& {{y}^{4}}-25{{y}^{2}}-16{{y}^{2}}+400=0 \\
& {{y}^{2}}\left( {{y}^{2}}-25 \right)-16\left( {{y}^{2}}-25 \right)=0 \\
& \left( {{y}^{2}}-25 \right)\left( {{y}^{2}}-16 \right)=0
\end{align}$
By using the algebraic identity $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right)$ to simplify the equation, we get:
$\begin{align}
& \left( {{y}^{2}}-25 \right)\left( {{y}^{2}}-16 \right)=0 \\
& \left( {{y}^{2}}-{{5}^{2}} \right)\left( {{y}^{2}}-{{4}^{2}} \right)=0 \\
& \left( y-5 \right)\left( y+5 \right)\left( y-4 \right)\left( y+4 \right)=0
\end{align}$
Then, the solution of the equation is $ y=5,y=-5$, $ y=4$, and $ y=-4$.
Put the value of y in the second equation to find the value of x.
For $ y=5$,
$\begin{align}
& xy=20 \\
& x\left( 5 \right)=20 \\
& x=\frac{20}{5} \\
& x=4
\end{align}$
For $ y=-5$,
$\begin{align}
& xy=20 \\
& x\left( -5 \right)=20 \\
& x=\frac{20}{-5} \\
& x=-4
\end{align}$
For $ y=4$,
$\begin{align}
& xy=20 \\
& x\left( 4 \right)=20 \\
& x=\frac{20}{4} \\
& x=5
\end{align}$
For $ y=-4$,
$\begin{align}
& xy=20 \\
& x\left( -4 \right)=20 \\
& x=\frac{20}{-4} \\
& x=-5
\end{align}$
The pairs of solutions for the system of equations is as shown below:
$\left\{ \left( 4,5 \right),\left( -4,-5 \right),\left( 5,4 \right),\left( -5,-4 \right) \right\}$.
Consider the sides of the rectangle passing through points $\left\{ \left( 4,5 \right),\left( -4,-5 \right),\left( 5,4 \right),\left( -5,-4 \right) \right\}$ is,
Length: $\sqrt{{{\left( 5-\left( -4 \right) \right)}^{2}}+{{\left( 4-\left( -5 \right) \right)}^{2}}}=9\sqrt{2}$
Width: $\sqrt{{{\left( 5-4 \right)}^{2}}+{{\left( 4-5 \right)}^{2}}}=\sqrt{2}$
Calculate the area of the rectangle my multiplying the length with the breadth as shown below:
$9\sqrt{2}\times \sqrt{2}=18$
Hence, the area of a rectangle is 18 square units.
