Chapter 11.5, Problem 21PPS

To determine

To find: The area of the given Figure and use the segment length to find the area of the similar polygon.

Answer to Problem 21PPS

The area of the triangle is $18$ and the area of the similar figure is $56.2$ .

Explanation of Solution

Given:

The given Figure is shown in Figure 1

  

Figure 1

The value of $B^{\prime }{C}^{\prime }=5$ .

Calculation:

Consider the area of the parallelogram is,

  $\begin{array}{c}A=\Delta ABC+\Delta ADC\\ =\frac{1}{2}bh+\frac{1}{2}bh\\ =\frac{1}{2}\left(6\right)\left(2\right)+\frac{1}{2}\left(6\right)\left(4\right)\\ =18\end{array}$

The length of $BC$ line by the formula of distance between two points is,

  $\begin{array}{c}BC=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =4+4\\ =\sqrt{8}\end{array}$

Since, $A^{\prime }{B}^{\prime }{C}^{\prime }{D}^{\prime }$ similar to $ABCD$ then scale factor is,

  $\begin{array}{c}k=\frac{BC}{B^{\prime }{C}^{\prime }}\\ =\frac{\sqrt{8}}{5}\\ =\frac{3}{4}\end{array}$

Then, according to theorem of similar polygon.

  $\begin{array}{l}{k}^2=\frac{18}{A}\\ {\left(\frac{\sqrt{8}}{5}\right)}^2=\frac{18}{A}\\ A=\frac{18\left(25\right)}{8}\\ A=56.2\end{array}$

Thus, the area of the triangle is $18$ and the area of the similar figure is $56.2$ .