Chapter 10, Problem 8RE

To determine

To find: The lengths of the sides of right triangle. Prove that these lengths satisfy Pythagorean Theorem.

Answer to Problem 8RE

The length of three sides of right triangle are $\sqrt{29}$ , $\sqrt{42}$ and $\sqrt{13}$ .

Explanation of Solution

Given information:

The graph is shown below.

  

Calculation:

The vertices of the right triangle are $\left(4,3,2\right)$ , $\left(0,0,4\right)$ and $\left(4,5,5\right)$ . The length of the sides of triangle is the distance between the vertices.

Use the distance formula for two points $\left(x_1,y_1,z_1\right)$ and $\left(x_2,y_2,z_2\right)$ .

  $\text{Distance}=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2}$

The distance between the points $\left(4,3,2\right)$ and $\left(0,0,4\right)$ is:

  $\begin{array}{c}\text{Distance}=\sqrt{{\left(0-4\right)}^2+{\left(0-3\right)}^2+{\left(4-2\right)}^2}\\ =\sqrt{{\left(-4\right)}^2+{\left(-3\right)}^2+{\left(2\right)}^2}\\ =\sqrt{16+9+4}\\ =\sqrt{29}\end{array}$

So, the length of perpendicular of triangle is $\sqrt{29}$ .

The distance between the points $\left(0,0,4\right)$ and $\left(4,5,5\right)$ is:

  $\begin{array}{c}\text{Distance}=\sqrt{{\left(4-0\right)}^2+{\left(5-0\right)}^2+{\left(5-4\right)}^2}\\ =\sqrt{{\left(4\right)}^2+{\left(5\right)}^2+{\left(1\right)}^2}\\ =\sqrt{16+25+1}\\ =\sqrt{42}\end{array}$

So, the length of hypotenuse of triangle is $\sqrt{42}$ .

The distance between the points $\left(4,3,2\right)$ and $\left(4,5,5\right)$ is:

  $\begin{array}{c}\text{Distance}=\sqrt{{\left(4-4\right)}^2+{\left(5-3\right)}^2+{\left(5-2\right)}^2}\\ =\sqrt{{\left(0\right)}^2+{\left(2\right)}^2+{\left(3\right)}^2}\\ =\sqrt{0+4+9}\\ =\sqrt{13}\end{array}$

So, the length of base of triangle is $\sqrt{13}$ .

Therefore, the length of three sides of right triangle are $\sqrt{29}$ , $\sqrt{42}$ and $\sqrt{13}$ .

Let us check the Pythagorean Theorem.

  ${\text{P}}^2+{\text{B}}^2={\text{H}}^2$

Take the left side of the equation.

  $\begin{array}{c}{\text{P}}^2+{\text{B}}^2={\left(\sqrt{29}\right)}^2+{\left(\sqrt{13}\right)}^2\\ =29+13\\ =42\\ ={\left(\sqrt{42}\right)}^2\end{array}$

Hence the lengths of the right triangle satisfy the Pythagorean Theorem.