A Norman window has the shape of a rectangle surmounted by a semicircle. If the perimeter of the window is 38 ft, express the area A of the window as a function of the width x of the window.

To express the area \( A \) of the Norman window as a function of the width \( x \), we need to break down the problem.

### Step 1: Geometry of the Window
- The window consists of a **rectangle** surmounted by a **semicircle**.
- Let the width of the window (and thus the width of the rectangle) be \( x \) feet.
- The height of the rectangular part is \( h \).
- The diameter of the semicircle is equal to the width of the window, i.e., \( x \).
- The radius of the semicircle is \( r = \frac{x}{2} \).

### Step 2: Perimeter of the Window
The perimeter consists of:
1. The two vertical sides of the rectangle (each of length \( h \)),
2. The width of the rectangle at the base (length \( x \)),
3. The curved part of the semicircle (circumference of a full circle is \( 2\pi r \), so for the semicircle it is \( \pi r = \frac{\pi x}{2} \)).

Thus, the perimeter of the window is:
\[
P = 2h + x + \frac{\pi x}{2}
\]
Given that the perimeter is 38 feet, we can set up the equation:
\[
2h + x + \frac{\pi x}{2} = 38
\]

### Step 3: Solving for \( h \)
Solving for \( h \) in terms of \( x \):
\[
2h = 38 - x - \frac{\pi x}{2}
\]
\[
h = \frac{38 - x - \frac{\pi x}{2}}{2}
\]
This is the height of the rectangular part in terms of \( x \).

### Step 4: Area of the Window
The total area of the window is the sum of the area of the rectangle and the area of the semicircle.
- The area of the rectangle is \( A_{\text{rect}} = xh \).
- The area of the semicircle is \( A_{\text{semi}} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi \left( \frac{x}{2} \right)^2 = \frac{\pi x^2}{8} \).

Thus, the total area is:
\[
A = A_{\text{rect}} + A_{\text{semi}} = xh + \frac{\pi x^2}{8}
\]
Substitute \( h \) from Step 3 into this equation:
\[
A = x \left( \frac{38 - x - \frac{\pi x}{2}}{2} \right) + \frac{\pi x^2}{8}
\]

This is the expression for the area \( A \) of the window as a function of the width \( x \).

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Would you like a step-by-step simplification of the area formula or further clarifications?

Here are 5 follow-up questions related to this concept:
1. How would you maximize the area of the window given the perimeter constraint?
2. What happens to the area as the width approaches the perimeter constraint?
3. How does the inclusion of the semicircle affect the total area compared to a simple rectangular window?
4. Can you graph the area function to visualize how the area changes with width?
5. How would the formula change if the top of the window were a full circle instead of a semicircle?

**Tip:** When solving geometry problems with constraints, always try to express unknowns in terms of a single variable to simplify solving and optimizing functions.

Explore how to express the area of a Norman window, consisting of a rectangle and a semicircle, as a function of its width given a perimeter constraint of 38 ft. Learn about key geometry and algebra concepts including perimeter and area formulas. Suitable for high school students in Grades 9-12.

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