Chapter 6.4, Problem 22E

a.

To determine

To complete: The given table.

Answer to Problem 22E

The table can be written as follows:

  

Explanation of Solution

Given information:

The following table:

  

The following similar figures:

  

To complete the table, find the ratio of lengths of corresponding sides and ratio of areas of $A$ to $B$ , $B$ to $C$ and $A$ to $C$ . To find ratio of side lengths, write a ratio comparing the lengths of a pair of corresponding sides. Then substitute the lengths of the sides and simplify. The ratio of lengths of corresponding sides of similar figures are equal.

Ratio of side lengths of $A$ to $B$ :

  $\begin{array}{l}\frac{\text{Width of }A}{\text{Width of }B}\text{ [Write ratio]}\\ \text{ =}\frac{1}{2}\text{ [Substitute values]}\\ \end{array}$

Ratio of areas of $A$ to $B$ :

  $\begin{array}{l}\frac{\text{Area of }A}{\text{Area of }B}\text{ [Write ratio]}\\ =\frac{1\cdot 2}{2\cdot 4}\text{ [Substitute values]}\\ \text{ =}\frac{1\cdot \cancel{2}}{\cancel{2}\cdot 4}\text{ [Divide out common factors]}\\ \text{ =}\frac{1}{4}\text{ [Simplify]}\end{array}$

Ratio of side lengths of $A$ to $C$ :

  $\begin{array}{l}\frac{\text{Width of }A}{\text{Width of }C}\text{ [Write ratio]}\\ \text{ =}\frac{1}{3}\text{ [Substitute values]}\\ \end{array}$

Ratio of areas of $A$ to $C$ :

  $\begin{array}{l}\frac{\text{Area of }A}{\text{Area of }C}\text{ [Write ratio]}\\ =\frac{1\cdot 2}{3\cdot 6}\text{ [Substitute values]}\\ \text{ =}\frac{1\cdot \cancel{2}}{3\cdot \cancel{\underset{3}{6}}}\text{ [Divide out common factors]}\\ \text{ =}\frac{1}{9}\text{ [Simplify]}\end{array}$

Ratio of side lengths of $B$ to $C$ :

  $\begin{array}{l}\frac{\text{Width of }B}{\text{Width of }C}\text{ [Write ratio]}\\ \text{ =}\frac{2}{3}\text{ [Substitute values]}\\ \end{array}$

Ratio of areas of $B$ to $C$ :

  $\begin{array}{l}\frac{\text{Area of }B}{\text{Area of }C}\text{ [Write ratio]}\\ =\frac{4\cdot 2}{3\cdot 6}\text{ [Substitute values]}\\ \text{ =}\frac{4\cdot \cancel{2}}{3\cdot \cancel{\underset{3}{6}}}\text{ [Divide out common factors]}\\ \text{ =}\frac{4}{9}\text{ [Simplify]}\end{array}$

The table can be written as follows:

  

b.

To determine

To explain: How the ratio of the areas of similar rectangles is related to the ratio of lengths of corresponding sides.

Answer to Problem 22E

The ratio of areas of similar rectangles is equal to the square of the ratio of their corresponding sides lengths.

Explanation of Solution

Given information:

The following similar figures:

  

The ratio of areas of similar rectangles is equal to the square of the ratio of their corresponding sides lengths.

  $\begin{array}{l}\text{Area of }A\text{ to }B={\left(\frac{1}{2}\right)}^2\text{=}\frac{1}{4}\\ \text{Area of }A\text{ to }C={\left(\frac{1}{3}\right)}^2\text{=}\frac{1}{9}\\ \text{Area of }B\text{ to }C={\left(\frac{2}{3}\right)}^2\text{=}\frac{4}{9}\end{array}$

c.

To determine

To calculate: Area of rectangle $D$ .

Answer to Problem 22E

Area of rectangle $D$ = $200$ units

Explanation of Solution

Given information:

The following similar figures:

  

Rectangle $D$ is similar to rectangle $A$ . Ratio of side length of rectangle $D$ to corresponding side length of rectangle $A$ is $10:1$ .

To find area of rectangle $D$ ,first find the ratio of area of $D\text{ to }A$ by squaring the ratio of given side length.

  $\text{Area of }D\text{ to }A={\left(\frac{10}{1}\right)}^2\text{=}\frac{100}{1}$

As calculated in subpart a, area of rectangle $A$ is $2$ units. Let Area of rectangle $D$ be $x$ units.

  $\begin{array}{l}\text{Area of }D\text{ to }A=\frac{\text{Area of }D}{\text{Area of }A}\\ \Rightarrow \frac{100}{1}=\frac{x}{2}\text{ [Substitute values]}\\ \Rightarrow x=2\cdot 100\text{ [Form cross product]}\\ \Rightarrow x=200\text{ [Multiply]}\end{array}$

Therefore, area of rectangle $D$ = $200$ units