Chapter 12.5, Problem 28E

To determine

To show: Whether $△ABC\cong △DEF$ .

Answer to Problem 28E

Yes, the triangles are not congruent.

Explanation of Solution

Given information:

  $A\left(-5,7\right),B\left(-5,2\right),C\left(0,2\right),D\left(0,6\right),E\left(0,1\right),F\left(4,1\right).$

Formula used:

Distance Formula:- $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Calculation:

Consider,

  $A\left(-5,7\right),B\left(-5,2\right),C\left(0,2\right),D\left(0,6\right),E\left(0,1\right),F\left(4,1\right).$

Measures of $△ABC$

Length of AB:

  $\begin{array}{l}A\left(-5,7\right),B\left(-5,2\right).\\ \left(x_1,y_1\right)=\left(-5,7\right),\\ \left(x_2,y_2\right)=\left(-5,2\right).\\ AB=\sqrt{{\left((-5)-(-5)\right)}^2+{\left(2-7\right)}^2}\end{array}$

  $\begin{array}{l}AB=\sqrt{{\left(-5+5\right)}^2+{\left(2-7\right)}^2}\\ AB=\sqrt{{\left(-5\right)}^2}\\ AB=5\end{array}$

Length of BC:

  $\begin{array}{l}B\left(-5,2\right),C\left(0,2\right).\\ \left(x_1,y_1\right)=\left(-5,2\right),\\ \left(x_2,y_2\right)=\left(0,2\right).\\ BC=\sqrt{{\left(0-(-5)\right)}^2+{\left(2-2\right)}^2}\end{array}$

  $\begin{array}{l}BC=\sqrt{{\left(0+5\right)}^2+{\left(2-2\right)}^2}\\ BC=\sqrt{{\left(5\right)}^2+{\left(0\right)}^2}\\ BC=5\end{array}$

Length of AC:

  $\begin{array}{l}A\left(-5,7\right),C\left(0,2\right).\\ \left(x_1,y_1\right)=\left(-5,7\right),\\ \left(x_2,y_2\right)=\left(0,2\right).\\ AC=\sqrt{{\left(0-(-5)\right)}^2+{\left(2-7\right)}^2}\end{array}$

  $\begin{array}{l}AC=\sqrt{{\left(0+5\right)}^2+{\left(2-7\right)}^2}\\ AC=\sqrt{{\left(5\right)}^2+{\left(-5\right)}^2}\\ AC=\sqrt{50}\end{array}$

Measures of $△DEF$

Length of DE:

  $\begin{array}{l}D\left(0,6\right),E\left(0,1\right).\\ \left(x_1,y_1\right)=\left(0,6\right),\\ \left(x_2,y_2\right)=\left(0,1\right).\\ DE=\sqrt{{\left(0-0\right)}^2+{\left(1-6\right)}^2}\end{array}$

  $\begin{array}{l}DE=\sqrt{{\left(0\right)}^2+{\left(-5\right)}^2}\\ DE=5\end{array}$

Length of EF:

  $\begin{array}{l}E\left(0,1\right),F\left(4,1\right).\\ \left(x_1,y_1\right)=\left(0,1\right),\\ \left(x_2,y_2\right)=\left(4,1\right).\\ EF=\sqrt{{\left(4-0\right)}^2+{\left(1-1\right)}^2}\end{array}$

  $\begin{array}{l}EF=\sqrt{{\left(4\right)}^2+{\left(0\right)}^2}\\ EF=4\end{array}$

Length of DF:

  $\begin{array}{l}D\left(0,6\right),F\left(4,1\right).\\ \left(x_1,y_1\right)=\left(0,6\right),\\ \left(x_2,y_2\right)=\left(4,1\right).\\ DF=\sqrt{{\left(4-0\right)}^2+{\left(1-6\right)}^2}\end{array}$

  $\begin{array}{l}DF=\sqrt{{\left(4\right)}^2+{\left(-5\right)}^2}\\ DF=\sqrt{41}\end{array}$

Since the sides of $△ABC\text{ and }△\text{DEF}$ are not equal.

Hence, triangles are not congruent.