Chapter 11, Problem 10RE

9.

To determine

Length of the sides of the triangle with vertices (0,5,-3), (0,3,2), (3,-2,0) and to prove the lengths satisfy Pythagorean theorem.

Answer to Problem 10RE

Lengths of the sides of the triangle are $\sqrt{29}$ units, $\sqrt{38}$ units and $\sqrt{67}$ units.

Explanation of Solution

Given:

The vertices of the triangle are (0,5,-3), (0,3,2), (3,-2,0).

Formula used:Distance between $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by $\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2+{(z_1-z_2)}^2}$ .

Calculation:

Length between the vertices (0,5,-3) and (0,3,2) is

  $\begin{array}{l}\sqrt{{(0-0)}^2+{(5-3)}^2+{(-3-2)}^2}\\ =\sqrt{0^2+2^2+{(-5)}^2}\\ =\sqrt{0+4+25}\\ =\sqrt{29}\end{array}$

Length between the vertices (0,3,2) and (3,-2,0) is

  $\begin{array}{l}\sqrt{{(0-3)}^2+{(3+2)}^2+{(2-0)}^2}\\ =\sqrt{{(-3)}^2+5^2+2^2}\\ =\sqrt{9+25+4}\\ =\sqrt{38}\end{array}$

Length between the vertices (3,-2,0) and (0,5,-3) is

  $\begin{array}{l}\sqrt{{(3-0)}^2+{(-2-5)}^2+{(0+3)}^2}\\ =\sqrt{3^2+{(-7)}^2+3^2}\\ =\sqrt{9+49+9}\\ =\sqrt{67}\end{array}$

Thus, the lengths of the three sides of the triangle are $\sqrt{29}$ units, $\sqrt{38}$ units and $\sqrt{67}$ units.

Pythagorean Theorem states that,

  ${(base)}^2+{(perpendicular)}^2={(hypotenuse)}^2$

Here,

  $\begin{array}{l}$ {(base)}^2+{(perpendicular)}^2$$ ={(\sqrt{38})}^2+{(\sqrt{29})}^2$=38+29\end{array}$

  $\begin{array}{l}=67\\ $ ={(\sqrt{67})}^2$$ ={(hypotenuse)}^2$\end{array}$

Hence proved, these lengths satisfy Pythagorean Theorem.

10.

To determine

Length of the sides of the triangle with vertices (0,0,4), (4,3,2), (4,5,5) and to prove the lengths satisfy Pythagorean theorem.

Answer to Problem 10RE

Lengths of the sides of the triangle are $\sqrt{29}$ units, $\sqrt{13}$ units and $\sqrt{42}$ units.

Explanation of Solution

Given:

The vertices of the triangle are (0,0,4), (4,3,2), (4,5,5).

Formula used:Distance between $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by $\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2+{(z_1-z_2)}^2}$ .

Calculation:

Length between the vertices (0,0,4) and (4,3,2) is

  $\begin{array}{l}\sqrt{{(0-4)}^2+{(0-3)}^2+{(4-2)}^2}\\ =\sqrt{{(-4)}^2+{(-3)}^2+2^2}\\ =\sqrt{16+9+4}\\ =\sqrt{29}\end{array}$

Length between the vertices (4,3,2) and (4,5,5) is

  $\begin{array}{l}\sqrt{{(4-4)}^2+{(3-5)}^2+{(2-5)}^2}\\ =\sqrt{0^2+{(-2)}^2+{(-3)}^2}\\ =\sqrt{0+4+9}\\ =\sqrt{13}\end{array}$

Length between the vertices (4,5,5) and (0,0,4) is

  $\begin{array}{l}\sqrt{{(4-0)}^2+{(5-0)}^2+{(5-4)}^2}\\ =\sqrt{4^2+5^2+1^2}\\ =\sqrt{16+25+1}\\ =\sqrt{42}\end{array}$

Thus, the lengths of the three sides of the triangle are $\sqrt{29}$ units, $\sqrt{13}$ units and $\sqrt{42}$ units.

Pythagorean Theorem states that,

  ${(base)}^2+{(perpendicular)}^2={(hypotenuse)}^2$

Here,

  $\begin{array}{l}$ {(base)}^2+{(perpendicular)}^2$$ ={(\sqrt{29})}^2+{(\sqrt{13})}^2$=29+13\end{array}$

  $\begin{array}{l}=42\\ $ ={(\sqrt{42})}^2$$ ={(hypotenuse)}^2$\end{array}$

Hence proved, these lengths satisfy Pythagorean Theorem.