https://r2.erweima.ai/mathbot/image/171860620580120eo1dl1.PNG

To solve the given problems in the image, let's break down the tasks:

1. **Derive an equation based on the areas in Figure 1:**
   - In Figure 1, a large square is formed using smaller rectangles and squares. Let’s denote the side length of the large square as \(a\) and \(b\), where \(a\) is the length of the rectangle and \(b\) is the width of the rectangle.
   - The area of the large square is \((a+b)^2\).
   - The large square is divided into four regions: two rectangles and two smaller squares.
     - The area of each smaller square is \(a^2\) and \(b^2\).
     - The area of each rectangle is \(ab\).

   Therefore, the equation based on the areas is:
   \[
   (a+b)^2 = a^2 + b^2 + 2ab
   \]

2. **Calculate the area of the shaded region in Figure 2 given \(a - b = 3\) and \(a^2 + b^2 = 15\):**
   - First, find the values of \(a\) and \(b\):
     - We have the equations:
       \[
       a - b = 3
       \]
       \[
       a^2 + b^2 = 15
       \]

     - Solve these equations simultaneously. From the first equation, express \(a\) in terms of \(b\):
       \[
       a = b + 3
       \]

     - Substitute this into the second equation:
       \[
       (b+3)^2 + b^2 = 15
       \]
       \[
       b^2 + 6b + 9 + b^2 = 15
       \]
       \[
       2b^2 + 6b + 9 = 15
       \]
       \[
       2b^2 + 6b - 6 = 0
       \]
       \[
       b^2 + 3b - 3 = 0
       \]

     - Solve this quadratic equation using the quadratic formula:
       \[
       b = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1}
       \]
       \[
       b = \frac{-3 \pm \sqrt{9 + 12}}{2}
       \]
       \[
       b = \frac{-3 \pm \sqrt{21}}{2}
       \]

       Therefore, \(b = \frac{-3 + \sqrt{21}}{2}\) (taking the positive root as length must be positive).

     - Now, find \(a\):
       \[
       a = b + 3 = \frac{-3 + \sqrt{21}}{2} + 3 = \frac{3 + \sqrt{21}}{2}
       \]

   - Calculate the area of the shaded region in Figure 2:
     - The shaded region is a right-angled triangle.
     - The base of the triangle is \(a\) and the height is \(b\).

     Therefore, the area of the shaded region is:
     \[
     \text{Area} = \frac{1}{2} \times a \times b = \frac{1}{2} \times \left(\frac{3 + \sqrt{21}}{2}\right) \times \left(\frac{-3 + \sqrt{21}}{2}\right)
     \]

     Simplify the expression:
     \[
     \text{Area} = \frac{1}{2} \times \frac{(3 + \sqrt{21})(-3 + \sqrt{21})}{4}
     \]
     \[
     \text{Area} = \frac{1}{2} \times \frac{-9 + 21}{4}
     \]
     \[
     \text{Area} = \frac{1}{2} \times \frac{12}{4}
     \]
     \[
     \text{Area} = \frac{1}{2} \times 3 = \frac{3}{2} = 1.5
     \]

Thus, the area of the shaded region in Figure 2 is \(1.5\) square units.

Do you want details or have any questions? Here are some more questions you might want to ask:

1. Can you provide more detailed steps for solving quadratic equations?
2. How can I verify the solution to the area problem?
3. Are there any alternative methods to solve this area problem?
4. Can you explain more about the properties of the shapes involved?
5. How would the problem change if \(a\) and \(b\) were different values?

**Tip:** Always verify your algebraic solutions by substituting back into the original equations to ensure consistency.

Explore how to solve geometric and algebraic problems illustrated in figures. This page covers topics such as the area of squares and rectangles, quadratic equations, and algebraic expressions. Suitable for students in Grades 10-12.

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