Chapter 6.5, Problem 52STP

a.

To determine

To calculate:

The area of $\mathrm{W}\mathrm{X}\mathrm{Y}\mathrm{Z}$

by using the following figure:

  

Answer to Problem 52STP

The area of $\mathrm{W}\mathrm{X}\mathrm{Y}\mathrm{Z}$ is $153\mathrm{c}{\mathrm{m}}^2$

Explanation of Solution

Given information:

  $\begin{array}{l}\mathrm{C}\mathrm{Y}=3\mathrm{c}\mathrm{m}\\ \mathrm{A}\left(\square \mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}\right)=81\mathrm{c}{\mathrm{m}}^2\end{array}$

Calculation:

The length of each side of $\square \mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$ is calculated as:

  $\begin{array}{l}\mathrm{A}\left(\square \mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}\right)={\mathrm{a}}^2\\ 81={\mathrm{a}}^2\\ \mathrm{a}=9\mathrm{c}\mathrm{m}\end{array}$

Now, $\mathrm{W}\mathrm{X}\mathrm{Y}\mathrm{Z}$ is formed by extending each side of $\square \mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$ by $3\mathrm{c}\mathrm{m}$ , so:

  $\mathrm{D}\mathrm{Z}=\mathrm{A}\mathrm{W}=\mathrm{B}\mathrm{X}=\mathrm{C}\mathrm{Y}=3\mathrm{c}\mathrm{m}$

Now, the area of triangle is calculated as:

  $\begin{array}{l}\mathrm{A}\left(\mathrm{\Delta }\right)=\text{}\frac{1}{2}\left(9+3\right)\left(3\right)\\ \mathrm{A}\left(\mathrm{\Delta }\right)=18\mathrm{c}{\mathrm{m}}^2\end{array}$

Therefore, the sum of four congruent triangles and area of $\square \mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$ is the area of $\square \mathrm{W}\mathrm{X}\mathrm{Y}\mathrm{Z}$ :

  $\begin{array}{l}\mathrm{A}\left(\square \mathrm{W}\mathrm{X}\mathrm{Y}\mathrm{Z}\right)=\mathrm{A}\left(\square \mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}\right)+\mathrm{A}\left(\mathrm{\Delta }\right)\\ \mathrm{A}\left(\square \mathrm{W}\mathrm{X}\mathrm{Y}\mathrm{Z}\right)=81+4\left(18\right)\\ \mathrm{A}\left(\square \mathrm{W}\mathrm{X}\mathrm{Y}\mathrm{Z}\right)=153\mathrm{c}{\mathrm{m}}^2\end{array}$

b.

To determine

To calculate:

The value of $\mathrm{D}\mathrm{Z}$

Answer to Problem 52STP

The value of $\mathrm{D}\mathrm{Z}$ is $5\mathrm{c}\mathrm{m}$

Explanation of Solution

Given information:

  $\begin{array}{l}\mathrm{A}\left(\square \mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}\right)=49\mathrm{c}{\mathrm{m}}^2\\ \mathrm{A}\left(\square \mathrm{W}\mathrm{X}\mathrm{Y}\mathrm{Z}\right)=169\mathrm{c}{\mathrm{m}}^2\end{array}$

Calculation:

Consider,

The length of segment $\overline{\mathrm{D}\mathrm{Z}}$ : $\mathrm{x}\mathrm{c}\mathrm{m}$

In the right angle triangle $\mathrm{W}\mathrm{Z}\mathrm{D}$ :

  $\begin{array}{l}\mathrm{W}\mathrm{Z}=13\mathrm{c}\mathrm{m}\\ \mathrm{W}\mathrm{D}=\mathrm{A}\mathrm{W}+\mathrm{A}\mathrm{D}=\mathrm{D}\mathrm{Z}=\mathrm{x}+7\end{array}$

Use the Pythagoras theorem to get value of $\mathrm{x}$ :

  $\begin{array}{l}{\mathrm{x}}^2+{\mathrm{x}}^2+14\mathrm{x}+49=169\\ 2{\mathrm{x}}^2+14\mathrm{x}-120=0\\ {\mathrm{x}}^2+7\mathrm{x}-60=0\\ \mathrm{x}=\frac{-7\pm \sqrt{7^2-4\left(1\right)\left(-60\right)}}{2\left(1\right)}\\ \mathrm{x}=\frac{-7\pm 17.3}{2}\\ \mathrm{x}=-12.5\mathrm{c}\mathrm{m}\mathrm{O}\mathrm{R}\text{}\mathrm{x}=5\mathrm{c}\mathrm{m}\end{array}$

The length is never negative so, $\mathrm{x}=5\mathrm{c}\mathrm{m}$ . Therefore, $\mathrm{D}\mathrm{Z}=5\mathrm{c}\mathrm{m}$

c.

To determine

To calculate:

The area of $\mathrm{W}\mathrm{X}\mathrm{Y}\mathrm{Z}$

by using the following figure:

  

Answer to Problem 52STP

The area of $\mathrm{W}\mathrm{X}\mathrm{Y}\mathrm{Z}$ is $2.5\mathrm{g}$ square meter

Explanation of Solution

Given information:

  $\begin{array}{l}\mathrm{A}\mathrm{B}=2\mathrm{C}\mathrm{Y}\\ \mathrm{A}\left(\square \mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}\right)=\mathrm{g}{\mathrm{m}}^2\end{array}$

Calculation:

If $\mathrm{A}\mathrm{B}=2\mathrm{C}\mathrm{Y}$ then the length of two legs of $\mathrm{\Delta }\mathrm{W}\mathrm{A}\mathrm{X}$ are:

  $\begin{array}{l}\mathrm{W}\mathrm{A}=\mathrm{C}\mathrm{Y}\\ \mathrm{A}\mathrm{X}=2\mathrm{C}\mathrm{Y}+\mathrm{C}\mathrm{Y}=3\mathrm{C}\mathrm{Y}\end{array}$

Use the Pythagoras theorem to get value of $\mathrm{x}$ :

  $\mathrm{W}{\mathrm{X}}^2=\mathrm{C}{\mathrm{Y}}^2+{\left(3\mathrm{C}\mathrm{Y}\right)}^2=10\mathrm{C}{\mathrm{Y}}^2$

Now,

  $10\mathrm{C}{\mathrm{Y}}^2=10{\left(\frac{\mathrm{A}\mathrm{B}}{2}\right)}^2=2.5\mathrm{A}{\mathrm{B}}^2$

But $\mathrm{A}\left(\square \mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}\right)=\mathrm{A}{\mathrm{B}}^2=\mathrm{g}$

Hence, the area of $\mathrm{W}\mathrm{X}\mathrm{Y}\mathrm{Z}$ is:

  $\mathrm{W}\mathrm{X}\mathrm{Y}\mathrm{Z}=\mathrm{W}{\mathrm{X}}^2=2.5\mathrm{g}{\mathrm{m}}^2$