Chapter 13.1, Problem 14PSB

(a)

To determine

To find: The length of the three sides of the triangle.

Answer to Problem 14PSB

The lengths of three sides are 3, 4, and 5.

Explanation of Solution

Given:

The vertices are $\left(0,0\right)$ , $\left(3,0\right)$ , and $\left(3,4\right)$ .

Calculation:

Use the distance formula to find the length of each sides.

Let the sides are $a$ , $b$ , and $c$ .

Find the length of side $a$ .

  $\begin{array}{c}a=\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}\\ =\sqrt{{\left(0-3\right)}^2+{\left(0-0\right)}^2}\\ =\sqrt{3^2}\\ =3\end{array}$

Find the length of side $b$ .

  $\begin{array}{c}b=\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}\\ =\sqrt{{\left(3-3\right)}^2+{\left(0-4\right)}^2}\\ =\sqrt{{\left(-4\right)}^2}\\ =4\end{array}$

Find the length of side $c$ .

  $\begin{array}{c}c=\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}\\ =\sqrt{{\left(3-0\right)}^2+{\left(4-0\right)}^2}\\ =\sqrt{9+16}\\ =5\end{array}$

Therefore, the lengths of three sides are 3, 4, and 5.

(b)

To determine

To find: The length of the altitude to the hypotenuse.

Answer to Problem 14PSB

The length of the altitude to the hypotenuse is $\frac{12}{5}$ .

Explanation of Solution

Given:

The vertices are $\left(0,0\right)$ , $\left(3,0\right)$ , and $\left(3,4\right)$ .

Calculation:

Consider the following figure.

  

Find the area of the triangle.

  $\begin{array}{c}A=\frac{1}{2}\left(4\right)\left(3\right)\\ =6\end{array}$

Let the length of the altitude to the hypotenuse be $h$ .

Find the length of the altitude to the hypotenuse.

  $\begin{array}{c}6=\frac{1}{2}\left(5h\right)\\ 6=\frac{5}{2}h\\ h=\frac{12}{5}\end{array}$

Therefore, the length of the altitude to the hypotenuse is $\frac{12}{5}$ .

(c)

To determine

To find: The length of the median to the hypotenuse.

Answer to Problem 14PSB

The length of the median to the hypotenuse is $2.5$ .

Explanation of Solution

Given:

The vertices are $\left(0,0\right)$ , $\left(3,0\right)$ , and $\left(3,4\right)$ .

Calculation:

Consider the following figure.

  

Find the mid-point of the line $\overleftrightarrow{\text{AC}}$ .

  $\begin{array}{c}\text{M}=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\ =\left(\frac{0+3}{2},\frac{10+4+3}{2}\right)\\ =\left(1.5,2\right)\end{array}$

Find the distance $\overleftrightarrow{\text{MB}}$ ^by using distance formula.

  $\begin{array}{c}d=\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}\\ =\sqrt{{\left(3-1.5\right)}^2+{\left(0-2\right)}^2}\\ =\sqrt{{1.5}^2+{\left(-2\right)}^2}\\ =2.5\end{array}$

Therefore, the length of the median to the hypotenuse is $2.5$ .