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

Let the original side - length of the square be $s$. The original perimeter $P_1 = 4s$ and the original area $A_1=s^{2}$.

Step2: Calculate new perimeter and area

When the side - length is tripled, the new side - length is $3s$. The new perimeter $P_2 = 4\times(3s)=12s$, and the new area $A_2=(3s)^{2}=9s^{2}$.

Step3: Calculate percent increase in perimeter

The percent increase in perimeter is $\frac{P_2 - P_1}{P_1}\times100\%=\frac{12s - 4s}{4s}\times100\%=\frac{8s}{4s}\times100\% = 200\%$.

Step4: Calculate percent increase in area

The percent increase in area is $\frac{A_2 - A_1}{A_1}\times100\%=\frac{9s^{2}-s^{2}}{s^{2}}\times100\%=\frac{8s^{2}}{s^{2}}\times100\% = 800\%$.

Step5: Compare percent increases

The percent increase in area is greater. The difference is $800\%−200\% = 600\%$.

Answer:

The percent of increase for the area is greater. It is 600% greater.