Question

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1. the side length of the square shown is tripled. which percent of increase is greater: the percent of increase for the perimeter of the square or the percent of increase for the area? how much greater?

Explanation:

Step1: Let the original side length of the square be \(a\)

• Original perimeter \(P_1 = 4a\)
• Original area \(A_1=a^{2}\)

Step2: Calculate the new side length, perimeter and area

• New side length \(a_2 = 3a\)
• New perimeter \(P_2=4\times(3a)=12a\)
• New area \(A_2=(3a)^{2}=9a^{2}\)

Step3: Calculate the percent increase for the perimeter

Percent increase formula: \(\text{Percent Increase}=\frac{\text{New Value}-\text{Original Value}}{\text{Original Value}}\times100\%\)
For perimeter: \(\frac{P_2 - P_1}{P_1}\times100\%=\frac{12a - 4a}{4a}\times100\%=\frac{8a}{4a}\times100\% = 200\%\)

Step4: Calculate the percent increase for the area

For area: \(\frac{A_2 - A_1}{A_1}\times100\%=\frac{9a^{2}-a^{2}}{a^{2}}\times100\%=\frac{8a^{2}}{a^{2}}\times100\%=800\%\)

Step5: Compare the two percent - increases

Subtract the percent increase of perimeter from the percent increase of area: \(800\% - 200\%=600\%\)

Answer:

The percent of increase for the area is greater. It is \(600\%\) greater than the percent of increase for the perimeter.