Chapter 4, Problem 24SGR

To determine

To explain:

  $△ABC\cong △XYZ$

Answer to Problem 24SGR

Congruence postulate shows $△ABC\cong △XYZ$

Explanation of Solution

Given information:

  $A\left(5,1\right),B\left(1,5\right),C\left(0,0\right),X\left(-3,3\right),Y\left(-7,6\right),Z\left(-8,1\right)$

If three sides of one triangle are congruent to here side of a second triangle, then the triangles are congruent.

find the length of $\overline{AB}$ with $A\left(5,2\right)B\left(1,5\right)$ .

  $\begin{array}{l}AB=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ AB=\sqrt{{\left(1-5\right)}^2+{\left(5-2\right)}^2}\\ AB=\sqrt{16+9}\\ AB=5\end{array}$

Find the length of $\overline{BC}$ with $B\left(1,5\right),C\left(0,0\right)$

  $\begin{array}{l}BC=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ BC=\sqrt{{\left(0-1\right)}^2+{\left(0-5\right)}^2}\\ BC=\sqrt{1+25}\\ BC=26\end{array}$

Find the length of $\overline{CA}$ with $C\left(0,0\right),A\left(5,2\right)$

  $\begin{array}{l}CA=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ CA=\sqrt{{\left(5-0\right)}^2+{\left(2-0\right)}^2}\\ CA=\sqrt{25+4}\\ CA=\sqrt{29}\end{array}$

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

Find the length of $\overline{XY}$ with $X\left(-3,3\right),Y\left(-7,6\right)$

  $\begin{array}{l}XY=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ XY=\sqrt{{\left(-7-\left(-3\right)\right)}^2+{\left(6-3\right)}^2}\\ XY=\sqrt{16+9}\\ XY=5\end{array}$

Find the length of $\overline{YZ}$ with $Y\left(-7,6\right),Z\left(-8,1\right)$

  $\begin{array}{l}YZ=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ YZ=\sqrt{{\left(-8-\left(-7\right)\right)}^2+{\left(1-6\right)}^2}\\ YZ=\sqrt{1+25}\\ YZ=\sqrt{26}\end{array}$

Find the length of $\overline{ZX}$ with $Z\left(-8,1\right),X\left(-3,3\right)$

  $\begin{array}{l}ZX=\sqrt{{\left(x_2-x_1\right)}^2+\left(y_2-y_1\right)}\\ ZX=\sqrt{{\left(-3-\left(-8\right)\right)}^2+{\left(3-1\right)}^2}\\ ZX=\sqrt{25+4}\\ ZX=\sqrt{29}\end{array}$

Thus, $\overline{AB}\cong \overline{XY},\overline{BC}\cong \overline{YZ},\overline{AC}\cong \overline{ZX}$ .