Question

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: Calculate original perimeter and area

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

The new side length is \(3s\).
The new perimeter \(P_2 = 4\times(3s)=12s\).
The new area \(A_2=(3s)^{2}=9s^{2}\).

Step3: Calculate percent increase for perimeter

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

Step4: Calculate percent increase for area

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

Step5: Compare the percent increases

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

Answer:

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