Chapter 4, Problem 17MCQ

To determine

To explain:

  $△PQR$ is congruent to $△XYZ$ or not.

Answer to Problem 17MCQ

Yes, $△PQR=△XYZ$ since corresponding sides are congruent.

Explanation of Solution

Given information:

  $P\left(3,-5\right),Q\left(11,0\right),R\left(1,6\right),X\left(5,1\right),Y\left(13,6\right),Z\left(3,12\right)$

find the length of each side of $△PQR$ .

Find the length of $\overline{PQ}$ with $P\left(3,-5\right),Q\left(11,0\right)$

  $\begin{array}{l}PQ=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(11-3\right)}^2+{\left(0-\left(-5\right)\right)}^2}\\ =\sqrt{64+25}\\ =\sqrt{89}\end{array}$

Find the length of $\overline{PR}$ with $P\left(3,-5\right),R\left(1,6\right)$

  $\begin{array}{l}PR=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ PR=\sqrt{{\left(1-3\right)}^2+{\left(6-\left(-5\right)\right)}^2}\\ PR=\sqrt{4+121}\\ PR=\sqrt{125}\\ BC=\sqrt{5}\end{array}$

Find the length of $\overline{QR}$ with $Q\left(11,0\right),R\left(1,6\right)$

  $\begin{array}{l}QR=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(1-11\right)}^2+{\left(6-0\right)}^2}\\ =\sqrt{100+36}\\ =\sqrt{136}\end{array}$

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

Find the length of $\overline{XY}$ with $X\left(5,1\right),Y\left(13,6\right)$

  $\begin{array}{l}XY=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ XY=\sqrt{{\left(13-5\right)}^2+{\left(6-1\right)}^2}\\ XY=\sqrt{64+25}\\ XY=\sqrt{89}\end{array}$

Find the length of $\overline{YZ}$ with $Y\left(13,6\right),Z\left(3,12\right)$

  $\begin{array}{l}YZ=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ YZ=\sqrt{{\left(3-13\right)}^2+{\left(12-6\right)}^2}\\ YZ=\sqrt{100+36}\\ YZ=\sqrt{136}\end{array}$

Find the length of $\overline{ZX}$ with $Z\left(3,12\right),X\left(5,1\right)$

  $\begin{array}{l}ZX=\sqrt{{\left(x_2-x_1\right)}^2+\left(y_2-y_1\right)}\\ ZX=\sqrt{{\left(5-3\right)}^2+{\left(1-12\right)}^2}\\ ZX=\sqrt{4+121}\\ ZX=\sqrt{125}\end{array}$

Thus, corresponding sides are congruent one obtains $△PQR=△XYZ$