Chapter 7.3, Problem 63E

To determine

To find the lengths of the sides of the triangle.

Answer to Problem 63E

The lengths of the sides of the triangles are $75\text{feet},42\text{feet},63\text{feet}$ .

And the longest side is $x=75$ feet, shortest is $z=63$ feet

Explanation of Solution

Given information:

The perimeter of a triangle is 180 feet.

The longest side of the triangle is 9 feet shorter than twice the shortest side.

The sum of the length of the two shorter sides is 30 feet more than the length of the longest side.

Calculation:

Let the sides of the triangle are x, y, z.

And x is a longest side and y is the shortest side.

Now, the perimeter of a triangle is the sum of the all sides.

  $x+y+z=180\mathrm{..........}\left(1\right)$

The longest side of the triangle is 9 feet shorter than twice the shortest side.

So, $x=2y-9\mathrm{...............}\left(2\right)$

The sum of the length of the two shorter sides is 30 feet more than the length of the longest side.

  $y+z=x+30$

Substitute $y+z=x+30$ in equation (1).

  $\begin{array}{c}x+y+z=180\mathrm{..........}\left(1\right)\\ x+x+30=180\\ 2x+30=180\\ 2x+30-30=180-30\left[\text{Subtract}\text{30}\text{from}\text{both}\text{sides}\right]\\ 2x=150\\ x=\frac{150}{2}\left[\text{divide}\text{by}\text{2}\right]\\ x=75\end{array}$

Substitute $x=75$ in equation (2).

  $\begin{array}{c}x=2y-9\mathrm{...............}\left(2\right)\\ 75=2y-9\\ 75+9=2y-9+9\left[\text{Add}\text{9}\text{from}\text{both}\text{sides}\right]\\ 84=2y\\ \frac{84}{2}=y\left[\text{divide}\text{by}\text{2}\right]\\ y=42\end{array}$

Substitute $x=75$ and $y=42$ in equation (1).

  $\begin{array}{c}x+y+z=180\mathrm{..........}\left(1\right)\\ 75+42+z=180\\ 117+z=180\\ 117-117+z=180-117\left[\text{subtract}\text{117}\text{from}\text{both}\text{sides}\right]\\ z=63\end{array}$

So, the lengths of the sides of the triangles are $75\text{feet},42\text{feet},63\text{feet}$ .

And the longest side is $x=75$ feet, shortest is $z=63$ feet