Chapter 11.1, Problem 35E

To determine

Tofind:The length of the sides of the right triangle with the given vertices and to show that these lengths satisfy the Pythagorean Theorem.

Answer to Problem 35E

The length of the sides of triangle are $\sqrt{20},3,\sqrt{29}$

Explanation of Solution

Given vertices: $\left(0,0,2\right),\left(0,4,0\right),\left(-2,5,2\right)$

  

Formula used: The distance between the points $\left(x_1,y_1,z_1\right)$ and $\left(x_2,y_2,z_2\right)$ is

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

Calculation:

Let ABC be the right triangle right angled at B with vertices $A\left(0,0,2\right),B\left(0,4,0\right),C\left(-2,5,2\right)$

Then by using distance formula , length of the sides of the triangle are

  $\begin{array}{l}AB=\sqrt{{\left(0-0\right)}^2+{\left(4-0\right)}^2+{\left(0-2\right)}^2}=\sqrt{0+16+4}=\sqrt{20}\\ BC=\sqrt{{\left(-2-0\right)}^2+{\left(5-4\right)}^2+{\left(2-0\right)}^2}=\sqrt{4+1+4}=3\\ CA=\sqrt{{\left(0+2\right)}^2+{\left(5-0\right)}^2+{\left(2-2\right)}^2}=\sqrt{4+25+0}=\sqrt{29}\end{array}$

Now,

  $\begin{array}{l}{\left(AB\right)}^2+{\left(BC\right)}^2={\left(\sqrt{20}\right)}^2+{\left(3\right)}^2=29\\ {\left(AB\right)}^2+{\left(BC\right)}^2={\left(CA\right)}^2\end{array}$

Therefore, these lengths satisfy the Pythagorean Theorem.