Question

Suppose the length and width of rectangle A are each 40% of the length and width of rectangle B. Is the area of rectangle A 40% of the area of rectangle B? Justify your answer.

Short answer

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

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

Step-by-step solution

Step 1. Given Information.

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

Step 2. Calculate the area of rectangle B.

Assume, length =$x$units

And width =$y$units

Therefore,

$\begin{array}{c}\text{area}=\text{length}\cdot \text{width}\\ =x\cdot y\\ =xy\text{unitsquare}\end{array}$

Step 3. Calculate the area of rectangle A.

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

$\begin{array}{c}=40\%\text{of}x\\ =\frac{40}{100}x\text{[writepercentasfraction]}\\ =\frac{2}{5}x\text{[simplify]}\end{array}$

Width = $40\%$ of the width of rectangleB.

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

Therefore,

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

Step 4. Conclusion.

Converting $\frac{4}{25}$ into percent

$\begin{array}{l}\frac{4}{25}=0.16\text{[convertfractionintodecimal]}\\ 0.16=16\%\text{[convertdecimalintopercentbyshiftingthedecimaltwoplaces}\\ \text{andwritingwithpercentsign]}\end{array}$

Therefore,

Area of rectangle A is $16\%$ of area of rectangle B.

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