Chapter 9.1, Problem 41PPS

(a)

To determine

To find: The sketch for triangle XYZ on coordinate plane.

Answer to Problem 41PPS

The sketch for triangle XYZ on coordinate plane is shown in figure 1.

Explanation of Solution

Given:

The coordinate of the triangle is given as,

  $X\left(4,9\right),Y\left(8,-9\right),Z\left(-6,5\right)$

Calculation:

The sketch for triangle XYZ on coordinate plane is shown in below figure.

  

Figure 1

Thus, the sketch for triangle XYZ on coordinate plane is shown in figure 1.

(b)

To determine

To find: The midpoint of the coordinates of triangle.

Answer to Problem 41PPS

The midpoint of the coordinates of triangle is (6, 0), (1, -2) and (-1, 7).

Explanation of Solution

Given:

The coordinate of the triangle is given as,

  $X\left(4,9\right),Y\left(8,-9\right),Z\left(-6,5\right)$

Calculation:

The coordinates of the mid-point for side XY is calculated as,

  $\begin{array}{c}\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)=\left(\frac{4+8}{2},\frac{9-9}{2}\right)\\ =\left(\frac{12}{2},\frac{0}{2}\right)\\ =\left(6,0\right)\end{array}$

The coordinates of the mid-point for side YZ is calculated as,

  $\begin{array}{c}\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)=\left(\frac{8-6}{2},\frac{-9+5}{2}\right)\\ =\left(\frac{2}{2},\frac{-4}{2}\right)\\ =\left(1,-2\right)\end{array}$

The coordinates of the mid-point for side ZX is calculated as,

  $\begin{array}{c}\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)=\left(\frac{-6+4}{2},\frac{5+9}{2}\right)\\ =\left(\frac{-2}{2},\frac{14}{2}\right)\\ =\left(-1,7\right)\end{array}$

Thus, the midpoint of the coordinates of triangle is (6, 0), (1, -2) and (-1, 7).

(c)

To determine

To find: The perimeter of XYZ and EFG

Answer to Problem 41PPS

The perimeter of XYZ is $\left(2\sqrt{85}+14\sqrt{2}+2\sqrt{29}\right)\text{units}$ and the perimeter of EFG is $\left(\sqrt{29}+\sqrt{85}+7\sqrt{2}\right)\text{ units}$

Explanation of Solution

Given:

The coordinate of the triangle is given as,

  $X\left(4,9\right),Y\left(8,-9\right),Z\left(-6,5\right)$

Calculation:

Calculate the distance XY using the distance formula.

  $\begin{array}{c}d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(8-4\right)}^2+{\left(-9-9\right)}^2}\\ =\sqrt{340}\\ =2\sqrt{85}\text{ units}\end{array}$

Calculate the distance YZ using the distance formula.

  $\begin{array}{c}d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(-6-8\right)}^2+{\left(5+9\right)}^2}\\ =\sqrt{392}\\ =14\sqrt{2}\text{ units}\end{array}$

Calculate the distance ZX using the distance formula.

  $\begin{array}{c}d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(-6-4\right)}^2+{\left(5-9\right)}^2}\\ =\sqrt{116}\\ =2\sqrt{29}\text{ units}\end{array}$

Calculate the distance EF using the distance formula.

  $\begin{array}{c}d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(1-6\right)}^2+{\left(-2-0\right)}^2}\\ =\sqrt{29}\text{ units}\end{array}$

Calculate the distance FG using the distance formula.

  $\begin{array}{c}d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(-1-1\right)}^2+{\left(7+2\right)}^2}\\ =\sqrt{85}\text{ units}\end{array}$

Calculate the distance ZX using the distance formula.

  $\begin{array}{c}d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(-1-6\right)}^2+{\left(7-0\right)}^2}\\ =\sqrt{98}\\ =7\sqrt{2}\text{ units}\end{array}$

Thus, the perimeter of XYZ is $\left(2\sqrt{85}+14\sqrt{2}+2\sqrt{29}\right)\text{units}$ and the perimeter of EFG is $\left(\sqrt{29}+\sqrt{85}+7\sqrt{2}\right)\text{ units}$

(d)

To determine

To find: The comparison of the perimeter with (c)

Answer to Problem 41PPS

The perimeter of triangle XYZ is twice of the perimeter of triangle EFG.

Explanation of Solution

Given:

The coordinate of the triangle is given as,

  $X\left(4,9\right),Y\left(8,-9\right),Z\left(-6,5\right)$

Calculation:

Calculate the perimeter.

  $\begin{array}{c}\text{Per}\left(\text{XYZ}\right)=\left(2\sqrt{85}+14\sqrt{2}+2\sqrt{29}\right)\text{units}\\ \text{=2}\left(\sqrt{85}+7\sqrt{2}+\sqrt{29}\right)\text{units}\\ =\text{2Per}\left(\text{EFG}\right)\end{array}$

From above, the perimeter of triangle XYZ is twice of the perimeter of triangle EFG.

Thus, the perimeter of triangle XYZ is twice of the perimeter of triangle EFG.