Chapter 7.2, Problem 39PPS

To determine

To check: Whether rectangle ABCD is similar to rectangle WXYZ.

Answer to Problem 39PPS

The rectangle ABCD is not similar to rectangle WXYZ.

Explanation of Solution

Given:

  $\begin{array}{l}A\left(5,5\right),B\left(0,0\right),C\left(5,-5\right),D\left(10,0\right)\\ W\left(1,6\right),X\left(-3,2\right),Y\left(2,-3\right),Z\left(6,1\right)\end{array}$

Calculation:

For rectangle ABCD,

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

For rectangle WXYZ,

  $\begin{array}{c}WX=\sqrt{{\left(-3-1\right)}^2+{\left(2-6\right)}^2}\\ =4\sqrt{2}\\ XY=\sqrt{{\left(2-\left(-3\right)\right)}^2+{\left(-3-2\right)}^2}\\ =5\sqrt{2}\end{array}$

Now, compare the lengths.

  $\begin{array}{c}\frac{AB}{WX}=\frac{5\sqrt{2}}{5\sqrt{2}}\\ =\frac{1}{1}\\ \frac{BC}{XY}=\frac{5\sqrt{2}}{4\sqrt{2}}\\ =\frac{5}{4}\end{array}$

Hence, both the rectangles are not similar.

Thus, the rectangle ABCD is not similar to rectangle WXYZ.