Chapter 10, Problem 7RE

To determine

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

Answer to Problem 7RE

The length of three sides of right triangle are $\sqrt{38}$ , $\sqrt{29}$ and $\sqrt{67}$ .

Explanation of Solution

Given information:

The graph is shown below.

  

Calculation:

The vertices of the right triangle are $\left(3,-2,0\right)$ , $\left(0,3,2\right)$ and $\left(0,5,-3\right)$ . 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 $\left(3,-2,0\right)$ and $\left(0,3,2\right)$ is:

  $\begin{array}{c}\text{Distance}==\sqrt{{\left(0-3\right)}^2+{\left(3-\left(-2\right)\right)}^2+{\left(2-0\right)}^2}\\ =\sqrt{{\left(-3\right)}^2+{\left(5\right)}^2+{\left(2\right)}^2}\\ =\sqrt{9+25+4}\\ =\sqrt{38}\end{array}$

So, the length of perpendicular of triangle is $\sqrt{38}$ .

The distance between the points $\left(0,3,2\right)$ and $\left(0,5,-3\right)$ is:

  $\begin{array}{c}\text{Distance}=\sqrt{{\left(0-0\right)}^2+{\left(5-3\right)}^2+{\left(-3-2\right)}^2}\\ =\sqrt{{\left(0\right)}^2+{\left(2\right)}^2+{\left(-5\right)}^2}\\ =\sqrt{0+4+25}\\ =\sqrt{29}\end{array}$

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

The distance between the points $\left(3,-2,0\right)$ and $\left(0,5,-3\right)$ is:

  $\begin{array}{c}\text{Distance}=\sqrt{{\left(0-3\right)}^2+{\left(5-\left(-2\right)\right)}^2+{\left(-3-0\right)}^2}\\ =\sqrt{{\left(-3\right)}^2+{\left(7\right)}^2+{\left(-3\right)}^2}\\ =\sqrt{9+49+9}\\ =\sqrt{67}\end{array}$

So, the length of hypotenuse of triangle is $\sqrt{67}$ .

Therefore, uhe length of three sides of right triangle are $\sqrt{38}$ , $\sqrt{29}$ and $\sqrt{67}$ .

Let us check the Pythagorean Theorem.

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

Take the left side of the equation.

  $\begin{array}{c}{\text{P}}^2+{\text{B}}^2={\left(\sqrt{38}\right)}^2+{\left(\sqrt{29}\right)}^2\\ =38+29\\ =67\\ ={\left(\sqrt{67}\right)}^2\end{array}$

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