Chapter 5.6, Problem 30E

a.

To determine

To find the area of the red rectangle.

Answer to Problem 30E

The area of the red rectangle is $\frac{5}{8}{\text{inch}}^2$

Explanation of Solution

Given information:

The figure shown is composed of two rectangles:

  

The area of the figure is $1\text{ square inch}$ .

Formula used: The following formula is used to calculate the area of the reactangle:

  $\text{Area of the rectangle}=\text{length}\times \text{breadth}$

Calculation :

The area of the red rectangle can be calculated by:

  $\begin{array}{l}\text{Area of the rectangle}=\text{length}\times \text{breadth}\\ \text{Area of the rectangle}=\left(1\times \frac{5}{8}\right){\text{inch}}^2\\ \text{Area of the rectangle}=\frac{5}{8}{\text{inch}}^2\end{array}$

Hence, the area of the red rectangle is $\frac{5}{8}{\text{inch}}^2$

b.

To determine

To write an expression for the area of blue rectangle.

Answer to Problem 30E

The expression for calculating the area of the blue rectangle is y $\left(\frac{3}{4}\times \text{x}\right){\text{ inch}}^2$ .

Explanation of Solution

Given information:

The figure shown is composed of two rectangles:

  

The area of the figure is 1 $1\text{ square inch}$ .

The area of the red rectangle is 3 $\frac{5}{8}{\text{inch}}^2$

Calculation :

The expression for calculating the area of the blue rectangle is:

  $\begin{array}{l}\text{Area of the blue rectangle}=\text{length}\times \text{breadth}\\ \text{Area of the blue rectangle}=\left(\frac{3}{4}\times \text{x}\right){\text{inch}}^2\\ \text{Area of the blue rectangle}=\frac{3}{4}{\text{x inch}}^2\end{array}$

Hence, the expression for calculating the area of the blue rectangle is $\left(\frac{3}{4}\times \text{x}\right){\text{ inch}}^2$ .

c.

To determine

To find the value of $\text{x}$ .

Answer to Problem 30E

The value of $\text{x}=\frac{1}{2}\text{inch}$ .

Explanation of Solution

Given information:

The figure shown is composed of two rectangles:

  

The area of the figure is $1\text{ square inch}$ .

The area of the red rectangle is $\frac{5}{8}{\text{inch}}^2$

Calculation :

The relation between the area of the red rectangle and the area of blue rectangle can be given by:

  $\begin{array}{l}\text{Area of the blue rectangle+Area of the red rectangle}=\text{Area of the figure}\\ \left(\text{length}\times \text{breadth}\right)+\left(\text{length}\times \text{breadth}\right)=\text{Area of the figure}\\ \left(\frac{3}{4}\times \text{x}\right){\text{inch}}^2+\left(\frac{5}{8}\times 1\right){\text{inch}}^2=1{\text{ inch}}^2\end{array}$

The value of $\text{x}$ can be calculated by:

  $\begin{array}{l}\left(\frac{3}{4}\times \text{x}\right){\text{inch}}^2+\left(\frac{5}{8}\times 1\right){\text{inch}}^2=1{\text{ inch}}^2\\ \left(\frac{3}{4}\text{x+}\frac{5}{8}\right)=1\\ \frac{6\text{x}+5}{8}=1\left[\text{multiply 8 to both sides}\right]\\ 6\text{x}+5=8\left[\text{subtracting 5 from both sides}\right]\\ 6\text{x}=3\left[\text{dividing both sides by 6}\right]\\ \text{x}=\frac{3}{6}\left[\text{Simplify}\right]\\ \text{x}=\frac{1}{2}\end{array}$

Hence, the value of $\text{x}=\frac{1}{2}\text{inch}$ .