Question

The hypotenuse of a right triangle is 8 feet long. One leg is 4 feet longer than the other. Find thelength of each leg to the nearest hundredth of a foot.

Answer 1

Given:

`\text{Hypotenuse}=8\text{feet}`

Let one leg be represented by x.

Hence,

`\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}`

The right triangle can be represented as shown below;

To solve for x, we use the Pythagoras theorem.

`\text{opposite}^2+adjacent^2=hypotenuse^2`

Hence,

`\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}`

Solving the quadratic equation by formula method,

`x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

For the equation,

`\begin{gathered} x^2+4x-24=0 \\ a=1 \\ b=4 \\ c=-24 \end{gathered}`

Substituting these values into the formula,

`\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x=\frac{-4\pm\sqrt[]{4^2-(4*1*-24)}}{2*1} \\ x=\frac{-4\pm\sqrt[]{16^{}-(-96)}}{2} \\ x=\frac{-4\pm\sqrt[]{16+96^{}}}{2} \\ x=\frac{-4\pm\sqrt[]{112}}{2} \\ x=(-4\pm10.58)/(2) \\ x_1=(-4+10.58)/(2)=(6.58)/(2)=3.29 \\ x_2=(-4-10.58)/(2)=(-14.58)/(2)=-7.29 \end{gathered}`

Since we are dealing with the length of a triangle, we discard the negative value.

Hence,

`x=3.29`

The length of one side is 3.29

The other side that is 4 feet longer will be,

`\begin{gathered} x+4=3.29+4 \\ =7.29 \end{gathered}`

Therefore, the length of each leg to the nearest hundredth of a foot is;

`\begin{gathered} 3.29\text{ f}eet \\ \\ \text{and} \\ \\ 7.29\text{ fe}et \end{gathered}`