The area of a rectangle is computed as follows:
A = length*height
In this case, the area is 54 ft², then:
54 = xy
Applying the Pythagorean theorem with the height, length, and diagonal of the rectangle we get:
[tex]\begin{gathered} c^2=a^2+b^2 \\ \sqrt[]{117}^2=x^2+y^2 \\ 117^{}=x^2+y^2 \end{gathered}[/tex]Isolating x from the first equation:
54/y = x
Substituting this result into the second equation:
[tex]\begin{gathered} 117=(\frac{54}{y})^2+y^2 \\ 117=\frac{2916}{y^2}^{}+y^2 \\ \text{Multiplying at both sides by y}^2 \\ 117y^2=\frac{2916}{y^2}y^2+y^2y^2 \\ 0=y^4-117y^2+2916 \end{gathered}[/tex]Replacing
[tex]\begin{gathered} t=y^2 \\ t^2=(y^2)^2=y^4 \\ 0=t^2-117t+2916 \end{gathered}[/tex]Applying the quadratic formula:
[tex]\begin{gathered} t_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ t_{1,2}=\frac{117\pm\sqrt[]{(-117)^2-4\cdot1\cdot2916}}{2\cdot1} \\ t_{1,2}=\frac{117\pm\sqrt[]{2025}}{2} \\ t_1=\frac{117+45}{2}=81 \\ t_2=\frac{117-45}{2}=36 \end{gathered}[/tex]Therefore, in terms of the original variable, y, the solutions are:
[tex]\begin{gathered} y^2_{}=t \\ y=\sqrt[]{t_1}=\sqrt[]{81}=\pm9 \\ y=\sqrt[]{t_2}=\sqrt[]{36}=\pm6 \end{gathered}[/tex]The negative results have no sense in this case, then the length and width are 9 ft and 6 ft, or vice versa.
