Chapter 11, Problem 9RE

To determine

To find: The length of the sides of given right angle triangle.

Answer to Problem 9RE

  $\sqrt{38},\sqrt{29},\text{ and }\sqrt{67}\text{ units}\text{.}$

Explanation of Solution

Given information:

A right angle triangle is given as below.

  

Calculation:

Distance between two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given as −

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

A right angle triangle is given. Label the three vertices as $P,Q,R$ .

  

Theside length $PQ$ will be given as −

  $\begin{array}{l}PQ=\sqrt{{(0-3)}^2+{(3+2)}^2+{(2-0)}^2}\\ PQ=\sqrt{9+25+4}\\ PQ=\sqrt{38}\end{array}$

  $\begin{array}{l}\text{Similarly, }\\ QR=\sqrt{{(0-0)}^2+{(5-3)}^2+{(-3-2)}^2}\\ QR=\sqrt{0+4+25}\\ QR=\sqrt{29}\\ \\ PR=\sqrt{{(0-3)}^2+{(5+2)}^2+{(-3-0)}^2}\\ PR=\sqrt{9+49+9}\\ PR=\sqrt{67}\end{array}$

Now using Pythagoras theorem,

  $\begin{array}{l}P{Q}^2+Q{R}^2=P{R}^2\\ \Rightarrow {\left(\sqrt{38}\right)}^2+{\left(\sqrt{29}\right)}^2={\left(\sqrt{67}\right)}^2\\ \Rightarrow 38+29=67\text{ This is true}\text{.}\end{array}$

Therefore, Pythagoras theorem is also satisfied.