Chapter 4.6, Problem 58SPR

To determine

To explain:

The triangles $\mathrm{\Delta }\mathrm{S}\mathrm{T}\mathrm{U}$ and $\mathrm{\Delta }\mathrm{X}\mathrm{Y}\mathrm{Z}$ are not congruentor not.

Answer to Problem 58SPR

The triangles $\mathrm{\Delta }\mathrm{S}\mathrm{T}\mathrm{U}$ and $\mathrm{\Delta }\mathrm{X}\mathrm{Y}\mathrm{Z}$ are not congruent.

Explanation of Solution

Given information:

  $\mathrm{S}\left(2,2\right),\mathrm{T}\left(4,6\right),\mathrm{U}\left(3,1\right),\mathrm{X}\left(-2,-2,\right),\mathrm{Y}\left(-4,-6\right),\mathrm{Z}\left(-3,1\right)$

Formula used:

The distance formula is given by the following expression:

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

Here, need to find the triangles are congruent or not. Use the distance formula to find the length of each side of the triangles: $\mathrm{S}\left(2,2\right),\mathrm{T}\left(4,6\right),\mathrm{U}\left(3,1\right),\mathrm{X}\left(-2,-2,\right),\mathrm{Y}\left(-4,-6\right),\mathrm{Z}\left(-3,1\right)$

In the triangle $\mathrm{\Delta }\mathrm{S}\mathrm{T}\mathrm{U}$ , the side lengths are calculated as:

  $\begin{array}{l}\mathrm{S}\mathrm{T}=\sqrt{{\left(4-2\right)}^2+{\left(6-2\right)}^2}=\sqrt{20}\\ \mathrm{T}\mathrm{U}=\sqrt{{\left(3-4\right)}^2+{\left(1-6\right)}^2}=\sqrt{26}\\ \mathrm{S}\mathrm{U}=\sqrt{{\left(4-2\right)}^2+{\left(6-2\right)}^2}=\sqrt{2}\end{array}$

In the triangle $\mathrm{\Delta }\mathrm{X}\mathrm{Y}\mathrm{Z}$ , the side lengths are calculated as:

  $\begin{array}{l}\mathrm{X}\mathrm{Y}=\sqrt{{\left(-4+2\right)}^2+{\left(6+2\right)}^2}=\sqrt{68}\\ \mathrm{Y}\mathrm{Z}=\sqrt{{\left(-3+4\right)}^2+{\left(1-6\right)}^2}=\sqrt{26}\\ \mathrm{Z}\mathrm{X}=\sqrt{{\left(-3+2\right)}^2+{\left(1+2\right)}^2}=\sqrt{10}\end{array}$

The corresponding sides are not congruent. Therefore, the triangles are not congruent.