Question
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
the percent increase for the (?) is greater.
it is (?)% greater
Answers
Answered by
GPT-4o mini
To compare the percent increase for the perimeter and the area of a square when the side length is increased, let's go through the calculations step by step.
1. **Perimeter of a Square**:
The perimeter \( P \) of a square with side length \( s \) is given by:
\[
P = 4s
\]
If the side length increases to \( s' = s + \Delta s \), the new perimeter \( P' \) is:
\[
P' = 4s' = 4(s + \Delta s) = 4s + 4\Delta s
\]
The increase in perimeter is:
\[
\Delta P = P' - P = (4s + 4\Delta s) - 4s = 4\Delta s
\]
The percent increase in perimeter is:
\[
\text{Percent Increase in Perimeter} = \left(\frac{\Delta P}{P}\right) \times 100 = \left(\frac{4\Delta s}{4s}\right) \times 100 = \left(\frac{\Delta s}{s}\right) \times 100
\]
2. **Area of a Square**:
The area \( A \) of a square is given by:
\[
A = s^2
\]
The new area \( A' \) after the increase in side length is:
\[
A' = (s')^2 = (s + \Delta s)^2 = s^2 + 2s\Delta s + (\Delta s)^2
\]
The increase in area is:
\[
\Delta A = A' - A = (s^2 + 2s\Delta s + (\Delta s)^2) - s^2 = 2s\Delta s + (\Delta s)^2
\]
The percent increase in area is:
\[
\text{Percent Increase in Area} = \left(\frac{\Delta A}{A}\right) \times 100 = \left(\frac{2s\Delta s + (\Delta s)^2}{s^2}\right) \times 100
\]
3. **Comparing the Percent Increases**:
- For the perimeter, the percent increase is:
\[
\left(\frac{\Delta s}{s}\right) \times 100
\]
- For the area (assuming \( \Delta s \) is much smaller than \( s \), making the \((\Delta s)^2\) term negligible), the percent increase simplifies to approximately:
\[
\left(\frac{2s\Delta s}{s^2}\right) \times 100 = \left(\frac{2\Delta s}{s}\right) \times 100
\]
4. **Conclusion**:
The percent increase for the area is approximately double that of the percent increase for the perimeter. Therefore:
The percent increase for the **area** is greater.
It is approximately **100%** greater than the percent increase for the perimeter.
1. **Perimeter of a Square**:
The perimeter \( P \) of a square with side length \( s \) is given by:
\[
P = 4s
\]
If the side length increases to \( s' = s + \Delta s \), the new perimeter \( P' \) is:
\[
P' = 4s' = 4(s + \Delta s) = 4s + 4\Delta s
\]
The increase in perimeter is:
\[
\Delta P = P' - P = (4s + 4\Delta s) - 4s = 4\Delta s
\]
The percent increase in perimeter is:
\[
\text{Percent Increase in Perimeter} = \left(\frac{\Delta P}{P}\right) \times 100 = \left(\frac{4\Delta s}{4s}\right) \times 100 = \left(\frac{\Delta s}{s}\right) \times 100
\]
2. **Area of a Square**:
The area \( A \) of a square is given by:
\[
A = s^2
\]
The new area \( A' \) after the increase in side length is:
\[
A' = (s')^2 = (s + \Delta s)^2 = s^2 + 2s\Delta s + (\Delta s)^2
\]
The increase in area is:
\[
\Delta A = A' - A = (s^2 + 2s\Delta s + (\Delta s)^2) - s^2 = 2s\Delta s + (\Delta s)^2
\]
The percent increase in area is:
\[
\text{Percent Increase in Area} = \left(\frac{\Delta A}{A}\right) \times 100 = \left(\frac{2s\Delta s + (\Delta s)^2}{s^2}\right) \times 100
\]
3. **Comparing the Percent Increases**:
- For the perimeter, the percent increase is:
\[
\left(\frac{\Delta s}{s}\right) \times 100
\]
- For the area (assuming \( \Delta s \) is much smaller than \( s \), making the \((\Delta s)^2\) term negligible), the percent increase simplifies to approximately:
\[
\left(\frac{2s\Delta s}{s^2}\right) \times 100 = \left(\frac{2\Delta s}{s}\right) \times 100
\]
4. **Conclusion**:
The percent increase for the area is approximately double that of the percent increase for the perimeter. Therefore:
The percent increase for the **area** is greater.
It is approximately **100%** greater than the percent increase for the perimeter.