Chapter 4, Problem 25SGR

To determine

To explain:

  $△ABC\cong △XYZ$

Answer to Problem 25SGR

Corresponding sides are not congruent one obtains $△ABC\ne △XYZ$

Explanation of Solution

Given information:

  $A\left(3,1\right),B\left(3,7\right),C\left(7,7\right),X\left(-7,0\right),Y\left(-7,4\right),Z\left(1,4\right)$

find he length of each side of $△ABC$ .

Find the length of $\overline{AB}$ with $A\left(3,1\right),B\left(3,7\right)$

  $\begin{array}{l}AB=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ AB=\sqrt{{\left(3-3\right)}^2+{\left(7-\left(-1\right)\right)}^2}\\ AB=\sqrt{0+64}\\ AB=\sqrt{8}\end{array}$

Find the length of $\overline{BC}$ with $B\left(3,7\right),C\left(7,7\right)$

  $\begin{array}{l}BC=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ BC=\sqrt{{\left(7-3\right)}^2+{\left(7-7\right)}^2}\\ BC=\sqrt{16+0}\\ BC=4\\ BC=\sqrt{5}\end{array}$

Find the length of $\overline{CA}$ with $C\left(7,7\right),A\left(3,-1\right)$

  $\begin{array}{l}CA=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ CA=\sqrt{{\left(3-7\right)}^2+{\left(-1-7\right)}^2}\\ CA=\sqrt{16+64}\\ CA=\sqrt{80}\end{array}$

Second find the length of each side $△XYZ$ ,

Find the length of $\overline{XY}$ with $X\left(-7,0\right),Y\left(-7,4\right)$

  $\begin{array}{l}XY=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ XY=\sqrt{{\left(-7-\left(-7\right)\right)}^2+{\left(4-0\right)}^2}\\ XY=\sqrt{0+16}\\ XY=4\end{array}$

Find the length of $\overline{YZ}$ with $Y\left(-7,4\right),Z\left(1,4\right)$

  $\begin{array}{l}YZ=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ YZ=\sqrt{{\left(1-\left(-7\right)\right)}^2+{\left(4-4\right)}^2}\\ YZ=\sqrt{64+0}\\ YZ=8\end{array}$

Find the length of $\overline{ZX}$ with $Z\left(1,4\right),X\left(-7,0\right)$

  $\begin{array}{l}ZX=\sqrt{{\left(x_2-x_1\right)}^2+\left(y_2-y_1\right)}\\ ZX=\sqrt{{\left(-7-1\right)}^2+{\left(0-4\right)}^2}\\ ZX=\sqrt{64+16}\\ ZX=\sqrt{80}\end{array}$

Thus, corresponding sides are not congruent one obtains $△ABC\ne △XYZ$