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The problem asks to express the total area \( A \) enclosed by two shapes (a square and a circle) formed by a 30-meter wire. The key to solving this problem is using the fact that the total length of the wire is split between the perimeter of the square and the circumference of the circle. Let's break down how to approach it:

### Given:
- Total length of wire = 30 meters.
- Part of the wire is shaped into a square with side length \( x \).
- The remaining part is shaped into a circle.

### Step-by-step solution:

1. **Wire for the square**:
   - The perimeter of the square is \( 4x \) since each side has length \( x \).
   - The wire used for the square is \( 4x \) meters.

2. **Wire for the circle**:
   - The remaining length of the wire is \( 30 - 4x \) meters, which forms the circumference of the circle.
   - The circumference of a circle is given by \( 2\pi r \), where \( r \) is the radius of the circle.
   - So, \( 2\pi r = 30 - 4x \), solving for \( r \), we get:
     \[
     r = \frac{30 - 4x}{2\pi}
     \]

3. **Area of the square**:
   - The area of the square is \( x^2 \).

4. **Area of the circle**:
   - The area of a circle is \( \pi r^2 \).
   - Using the expression for \( r \) from earlier, the area of the circle becomes:
     \[
     \text{Area of circle} = \pi \left(\frac{30 - 4x}{2\pi}\right)^2
     \]
     Simplifying:
     \[
     \text{Area of circle} = \frac{(30 - 4x)^2}{4\pi}
     \]

5. **Total area**:
   - The total area \( A(x) \) is the sum of the area of the square and the area of the circle:
     \[
     A(x) = x^2 + \frac{(30 - 4x)^2}{4\pi}
     \]

This expression gives the total area as a function of \( x \), the side length of the square.

Would you like more details on any step or a breakdown of how to simplify further?

Here are 5 questions to explore this further:
1. What is the domain of \( x \) in this problem?
2. How does the value of \( A(x) \) change as \( x \) increases?
3. What happens to the area of the circle if the entire wire is used for the square?
4. What are the units of \( A(x) \)?
5. How would you solve for the maximum area enclosed?

**Tip**: Always check the physical constraints of your variables, especially when dealing with lengths and areas!

A wire 30 meters long is to be cut into two pieces. One piece will be shaped as a square, and the other piece will be shaped as a circle. Express the total area enclosed by the pieces of wire as a function of the length x of a side of the square.

Learn how to express the total area of two shapes—a square and a circle—formed by a 30-meter wire. This problem involves key geometry concepts such as perimeter, circumference, and area formulas. Suitable for students in Grades 10-12.

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