Solution:
Given:
[tex]\text{Hypotenuse}=8\text{feet}[/tex]Let one leg be represented by x.
Hence,
[tex]\begin{gathered} \text{One leg= x} \\ \text{The other leg is 4 f}eet\text{ longer than the other,} \\ \text{Thus the other leg will be; x + 4} \end{gathered}[/tex]The right triangle can be represented as shown below;
To solve for x, we use the Pythagoras theorem.
[tex]\text{opposite}^2+adjacent^2=hypotenuse^2[/tex]Hence,
[tex]\begin{gathered} x^2+(x+4)^2=8^2 \\ x^2+(x+4)(x+4)=64 \\ x^2+x^2+4x+4x+16=64 \\ 2x^2+8x+16=64 \\ \text{Collecting all terms to one side to form a quadratic equation;} \\ 2x^2+8x+16-64=0 \\ 2x^2+8x-48=0_{} \\ \\ \text{Dividing the equation all through by 2,} \\ x^2+4x-24=0 \end{gathered}[/tex]Solving the quadratic equation by formula method,
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]For the equation,
[tex]\begin{gathered} x^2+4x-24=0 \\ a=1 \\ b=4 \\ c=-24 \end{gathered}[/tex]Substituting these values into the formula,
[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x=\frac{-4\pm\sqrt[]{4^2-(4\times1\times-24)}}{2\times1} \\ x=\frac{-4\pm\sqrt[]{16^{}-(-96)}}{2} \\ x=\frac{-4\pm\sqrt[]{16+96^{}}}{2} \\ x=\frac{-4\pm\sqrt[]{112}}{2} \\ x=\frac{-4\pm10.58}{2} \\ x_1=\frac{-4+10.58}{2}=\frac{6.58}{2}=3.29 \\ x_2=\frac{-4-10.58}{2}=\frac{-14.58}{2}=-7.29 \end{gathered}[/tex]Since we are dealing with the length of a triangle, we discard the negative value.
Hence,
[tex]x=3.29[/tex]The length of one side is 3.29
The other side that is 4 feet longer will be,
[tex]\begin{gathered} x+4=3.29+4 \\ =7.29 \end{gathered}[/tex]Therefore, the length of each leg to the nearest hundredth of a foot is;
[tex]\begin{gathered} 3.29\text{ f}eet \\ \\ \text{and} \\ \\ 7.29\text{ fe}et \end{gathered}[/tex]
