Chapter 4.7, Problem 20PPS

To determine

To graph:

Each pair of triangles with the following vertices and tell that the type of transformation by verification of the congruence transformation:

  $\begin{array}{l}\mathrm{A}\left(2,2\right),\mathrm{B}\left(4,7\right),\mathrm{C}\left(6,2\right);\\ \mathrm{D}\left(2,-2\right),\mathrm{F}\left(4,-7\right),\mathrm{G}\left(6,-2\right)\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}\mathrm{A}\left(2,2\right),\mathrm{B}\left(4,7\right),\mathrm{C}\left(6,2\right);\\ \mathrm{D}\left(2,-2\right),\mathrm{F}\left(4,-7\right),\mathrm{G}\left(6,-2\right)\end{array}$

Graph:

  

Interpretation:

From the drawn graph of the triangle with the given vertices, understands that each vertex of triangle and its image are the same distance from the $\mathrm{y}$ -axis.

Therefore, it is reflection which implies that $\mathrm{\Delta }\mathrm{A}\mathrm{B}\mathrm{C}$ is reflection of $\mathrm{\Delta }\mathrm{D}\mathrm{F}\mathrm{G}$ . Use the distance formula to find the length of each side of the triangles $\mathrm{\Delta }\mathrm{A}\mathrm{B}\mathrm{C}$ and $\mathrm{\Delta }\mathrm{D}\mathrm{F}\mathrm{G}$

The distance formula is given by the following expression:

  $\mathrm{d}=\sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2}$

In the triangle $\mathrm{\Delta }\mathrm{A}\mathrm{B}\mathrm{C}$ , the side lengths are calculated as:

  $\begin{array}{l}\overline{\mathrm{A}\mathrm{B}}=\sqrt{{\left(4-2\right)}^2+{\left(7-2\right)}^2}=\sqrt{29}\\ \overline{\mathrm{B}\mathrm{C}}=\sqrt{{\left(6-4\right)}^2+{\left(2-7\right)}^2}=\sqrt{29}\\ \overline{\mathrm{A}\mathrm{C}}=\sqrt{{\left(2-6\right)}^2+{\left(2-2\right)}^2}=4\end{array}$

In the triangle $\mathrm{\Delta }\mathrm{D}\mathrm{F}\mathrm{G}$ , the side lengths are calculated as:

  $\begin{array}{l}\overline{\mathrm{D}\mathrm{F}}=\sqrt{{\left(4-2\right)}^2+{\left(-7+2\right)}^2}=\sqrt{29}\\ \overline{\mathrm{F}\mathrm{G}}=\sqrt{{\left(6-4\right)}^2+{\left(-2+7\right)}^2}=\sqrt{29}\\ \overline{\mathrm{D}\mathrm{G}}=\sqrt{{\left(2-6\right)}^2+{\left(-2+2\right)}^2}=4\end{array}$

From the above calculations:

  $\begin{array}{l}\overline{\mathrm{A}\mathrm{B}}\cong \overline{\mathrm{D}\mathrm{F}}\\ \overline{\mathrm{B}\mathrm{C}}\cong \overline{\mathrm{F}\mathrm{G}}\\ \overline{\mathrm{A}\mathrm{C}}\cong \overline{\mathrm{D}\mathrm{G}}\end{array}$

Therefore, each pair of corresponding sides has the same measure that means they are congruent by SSS theorem.

Therefore,

  $\mathrm{\Delta }\mathrm{A}\mathrm{B}\mathrm{C}\cong \mathrm{\Delta }\mathrm{D}\mathrm{F}\mathrm{G}$