Chapter 11, Problem 4CLT

To determine

To prove: Compute the lengths of the triangle and show that they satisfy the Pythagorean Theorem.

Answer to Problem 4CLT

The given triangle is satisfying the Pythagorean Theorem and the lengths are 3 units, 4 units, and 5 units.

Explanation of Solution

Given information: The vertices of the triangle are (0,0,3), ( 0,4,3), and (0,0,0).

Let, triangle ABO

Where, AB is perpendicular, AO is base, and AO is hypotenuse

  

Formula used: Distance between two points, $\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2}$

Calculation:

Calculate the length of AB:

  $\begin{array}{l}\therefore AB=\sqrt{{\left(0-0\right)}^2+{\left(4-0\right)}^2+{\left(3-3\right)}^2}\\ =\sqrt{0+16+0}\\ =\sqrt{16}\\ =4\text{ units}\\ \therefore A{B}^2=16\end{array}$

Calculate the length of AO:

  $\begin{array}{l}\therefore AO=\sqrt{{\left(0-0\right)}^2+{\left(0-0\right)}^2+{\left(0-3\right)}^2}\\ =\sqrt{0+0+9}\\ =\sqrt{9}\\ =3\text{ units}\\ \therefore A{O}^2=9\end{array}$

Calculate the length of BO:

  $\begin{array}{l}\therefore BO=\sqrt{{\left(0-0\right)}^2+{\left(4-0\right)}^2+{\left(3-0\right)}^2}\\ =\sqrt{0+16+9}\\ =\sqrt{25}\\ =5\text{ units}\\ \therefore B{O}^2=25\end{array}$

As we know, according to Pythagorean Theorem

  $Hypotenus{e}^2 =Perpendicula{r}^2 +Bas{e}^2$ Or $B{O}^2 =A{B}^2 +A{O}^2$

Now, put the value of AB, AO, and BO

  $\begin{array}{l}\therefore 25=\text{ 16+9}\\ \text{or, 25=25}\end{array}$

Hence, the lengths are 3 units, 4 units, and the given triangle is right angle at angle A.