Chapter 7.1, Problem 58E

To determine

To find:The area of rectangle A is $40\%$ of the area of rectangle B or not.

Answer to Problem 58E

Area of rectangle A is not $40\%$ of the area of rectangle B.

Because by calculation,Area of rectangle A is $\text{16\% }$ of area of rectangle B.

Explanation of Solution

Given information:

The given information:

Length and width of rectangle A are each $40\%$ of the length and width of rectangle B.

As per the question,

For rectangle B.

Assume, length= $x$ units

And width= $y$ units

Therefore,

  $\begin{array}{l}area=length\cdot width\\ =x\cdot y\\ =xy\text{ unit square}\text{.}\end{array}$

For rectangle A,

Length = $40\%$ of the length of rectangle B.

  $\begin{array}{l}=40\%\text{ of }x\\ \text{=}\frac{40}{100}x\text{ [write percent as fraction]}\\ \text{=}\frac{2}{5}x\text{ [simplify]}\end{array}$

Width = $40\%$ of the width of rectangle B.

  $\begin{array}{l}=40\%\text{ of }y\\ \text{=}\frac{40}{100}y\text{ [write percent as fraction]}\\ \text{=}\frac{2}{5}y\text{ [simplify]}\end{array}$

Therefore,

  $\begin{array}{l}area=length\cdot width\\ =\frac{2}{5}x\cdot \frac{2}{5}y\\ \text{ =}\frac{4}{25}xy\text{ unit square}\text{. [multiply]}\end{array}$

Converting

  $\frac{4}{25}$ into percent

  $\begin{array}{l}\frac{4}{25}=0.16\text{ [convert fraction into decimal]}\\ \text{0}\text{.16=16\% [convert decimal into percent by shifting the decimal two places and writing with percent sign]}\end{array}$

Therefore,

Area of rectangle A= $\text{16\% }$ of area of rectangle B.

Hence,

Area of rectangle A is not $40\%$ of the area of rectangle B.