1. **Building Dimensions:** One World Trade Center is 1776 feet tall with a 200 ft by 200 ft square footprint.
2. **Walkway:** The building has a walkway around it, with a width denoted by \( x \).
3. **Equation:** The total area \( A \) is given by \( A = (200 + 2x)^2 \).
4. **Calculations:** If \( x = 13 \), \( A = 51,076 \) sq ft. If \( A = 75,000 \), \( x \approx 23.79 \) ft.
a. **Diagram:**
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b. **Equation:**
Let \( S \) be the width of the walkway.
The total area \( A \) is the sum of the area of the building and the area surrounding it.
\[ A = (200 + 2S)^2 \]
c. **If \( S = 13 \) feet, what is the total area?**
Substitute \( S = 13 \) into the equation:
\[ A = (200 + 2(13))^2 = 226^2 = 51,076 \, \text{square feet} \]
d. **If the total area is 75,000 square feet, what is the width of the walkway?**
Set \( A \) to 75,000 and solve for \( S \):
\[ 75,000 = (200 + 2S)^2 \]
\[ \sqrt{75,000} = 200 + 2S \]
\[ \sqrt{75,000} - 200 = 2S \]
\[ S = \frac{\sqrt{75,000} - 200}{2} \]
\[ S \approx 23.79 \]
So, if the total area is 75,000 square feet, the width of the walkway is approximately 23.79 feet.