Question

The side length of the square shown is tripled. (3 cm)
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?

the percent increase for the (?) is greater.

it is (?)% greater

Answer 1

To find the percent increase for both the perimeter and the area when the side length of a square is tripled, we start with the original side length of the square, which is 3 cm.

**1. Calculate the original perimeter and area:**

- The original side length: \(s = 3 \, \text{cm}\)
- Original perimeter \(P\):
\[
P = 4s = 4 \times 3 = 12 \, \text{cm}
\]
- Original area \(A\):
\[
A = s^2 = 3^2 = 9 \, \text{cm}^2
\]

**2. Calculate the new side length, perimeter, and area when the side length is tripled:**

- New side length:
\[
s' = 3s = 3 \times 3 = 9 \, \text{cm}
\]
- New perimeter \(P'\):
\[
P' = 4s' = 4 \times 9 = 36 \, \text{cm}
\]
- New area \(A'\):
\[
A' = (s')^2 = 9^2 = 81 \, \text{cm}^2
\]

**3. Calculate the percent increase for perimeter and area:**

- Percent increase for perimeter:
\[
\text{Percent Increase in Perimeter} = \frac{P' - P}{P} \times 100 = \frac{36 - 12}{12} \times 100 = \frac{24}{12} \times 100 = 200\%
\]

- Percent increase for area:
\[
\text{Percent Increase in Area} = \frac{A' - A}{A} \times 100 = \frac{81 - 9}{9} \times 100 = \frac{72}{9} \times 100 = 800\%
\]

**4. Comparison of percent increases:**

The percent increase for the area is \(800\%\) and for the perimeter is \(200\%\).

**Conclusion:**

The percent increase for the area is greater.

**Final Answer:**

The percent increase for the area is greater.

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