Chapter 12.5, Problem 25E

To determine

To show: Whether $△ABC\cong △DEF$ .

Answer to Problem 25E

Yes, the triangles are congruent.

Explanation of Solution

Given information:

  $A\left(-2,-2\right),B\left(4,-2\right),C\left(4,6\right),D\left(5,7\right),E\left(5,1\right),F\left(13,1\right).$

Formula used:

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

Calculation:

Consider,

  $A\left(-2,-2\right),B\left(4,-2\right),C\left(4,6\right),D\left(5,7\right),E\left(5,1\right),F\left(13,1\right).$

Measures of $△ABC$

Length of AB:

  $\begin{array}{l}A\left(-2,-2\right),B\left(4,-2\right).\\ \left(x_1,y_1\right)=\left(-2,-2\right),\\ \left(x_2,y_2\right)=\left(4,-2\right).\\ AB=\sqrt{{\left(4-(-2)\right)}^2+{\left((-2)-(-2)\right)}^2}\end{array}$

  $\begin{array}{l}AB=\sqrt{{\left(4+2\right)}^2+{\left(-2+2\right)}^2}\\ AB=\sqrt{{\left(6\right)}^2}\\ AB=6\end{array}$

Length of BC:

  $\begin{array}{l}B\left(4,-2\right),C\left(4,6\right).\\ \left(x_1,y_1\right)=\left(4,-2\right),\\ \left(x_2,y_2\right)=\left(4,6\right).\\ BC=\sqrt{{\left(4-4\right)}^2+{\left(6-(-2)\right)}^2}\end{array}$

  $\begin{array}{l}BC=\sqrt{{\left(4-4\right)}^2+{\left(6+2\right)}^2}\\ BC=\sqrt{{\left(8\right)}^2}\\ BC=8\end{array}$

Length of AC:

  $\begin{array}{l}A\left(-2,-2\right),C\left(4,6\right).\\ \left(x_1,y_1\right)=\left(-2,-2\right),\\ \left(x_2,y_2\right)=\left(4,6\right).\\ AC=\sqrt{{\left(4-(-2)\right)}^2+{\left(6-(-2)\right)}^2}\end{array}$

  $\begin{array}{l}AC=\sqrt{{\left(4+2\right)}^2+{\left(6+2\right)}^2}\\ AC=\sqrt{{\left(6\right)}^2+{\left(8\right)}^2}\\ AC=\sqrt{100}\\ AC=50\end{array}$

Measures of $△DEF$

Length of DE:

  $\begin{array}{l}D\left(5,7\right),E\left(5,1\right).\\ \left(x_1,y_1\right)=\left(5,7\right),\\ \left(x_2,y_2\right)=\left(5,1\right).\\ DE=\sqrt{{\left(5-5\right)}^2+{\left(1-7\right)}^2}\end{array}$

  $\begin{array}{l}DE=\sqrt{{\left(0\right)}^2+{\left(-6\right)}^2}\\ DE=\sqrt{{\left(-6\right)}^2}\\ DE=6\end{array}$

Length of EF:

  $\begin{array}{l}E\left(5,1\right),F\left(13,1\right).\\ \left(x_1,y_1\right)=\left(5,1\right),\\ \left(x_2,y_2\right)=\left(13,1\right).\\ EF=\sqrt{{\left(13-5\right)}^2+{\left(1-1\right)}^2}\end{array}$

  $\begin{array}{l}EF=\sqrt{{\left(8\right)}^2+{\left(0\right)}^2}\\ EF=\sqrt{{\left(8\right)}^2}\\ EF=8\end{array}$

Length of DF:

  $\begin{array}{l}D\left(5,7\right),F\left(13,1\right).\\ \left(x_1,y_1\right)=\left(5,7\right),\\ \left(x_2,y_2\right)=\left(13,1\right).\\ DF=\sqrt{{\left(13-5\right)}^2+{\left(1-7\right)}^2}\end{array}$

  $\begin{array}{l}DF=\sqrt{{\left(8\right)}^2+{\left(-6\right)}^2}\\ DF=\sqrt{{\left(8\right)}^2+{\left(6\right)}^2}\\ DF=\sqrt{100}\\ DF=50\end{array}$

Since the sides of $△ABC\text{ and }△\text{DEF}$ are equal.

By SSS Congruence Theorem,

  $△ABC\cong △DEF$

Hence, triangles are congruent.