https://r2.erweima.ai/mathbot/image/1718332983974r0f298jx.png

The problem in the image involves finding the area of the shaded region in a unit square \(ABCD\) where \(BE = 2EC\) and \(F\) is the midpoint of \(CD\).

Let's break down the problem step-by-step.

1. **Identify and label the points:**
   - \(A\) is at \((0, 1)\)
   - \(B\) is at \((0, 0)\)
   - \(C\) is at \((1, 0)\)
   - \(D\) is at \((1, 1)\)
   - \(F\) is the midpoint of \(CD\), so \(F\) is at \((1, 0.5)\)
   - \(BE = 2EC\), which means \(E\) divides \(BC\) in the ratio 2:1. Given that \(B\) is at \((0, 0)\) and \(C\) is at \((1, 0)\), \(E\) is at \((\frac{2}{3}, 0)\)

2. **Find the coordinates of \(G\), the intersection point of \(AF\) and \(DE\):**

   To find \(G\), we need the equations of lines \(AF\) and \(DE\):

   - **Equation of \(AF\):**
     The line \(AF\) passes through \(A(0, 1)\) and \(F(1, 0.5)\).
     The slope \(m_{AF}\) is \((0.5 - 1) / (1 - 0) = -0.5\).
     Using the point-slope form, the equation is \(y = -0.5x + 1\).

   - **Equation of \(DE\):**
     The line \(DE\) passes through \(D(1, 1)\) and \(E(\frac{2}{3}, 0)\).
     The slope \(m_{DE}\) is \((0 - 1) / (\frac{2}{3} - 1) = -3\).
     Using the point-slope form, the equation is \(y = -3x + 4\).

   Setting the equations equal to each other to find the intersection:
   \[
   -0.5x + 1 = -3x + 4
   \]
   Solving for \(x\):
   \[
   4 - 1 = 3x - 0.5x \\
   3 = 2.5x \\
   x = \frac{6}{5}
   \]

   Substituting \(x = \frac{6}{5}\) back into \(y = -0.5x + 1\):
   \[
   y = -0.5 \times \frac{6}{5} + 1 = 1 - 0.6 = 0.4
   \]

   Therefore, the coordinates of \(G\) are \(\left(\frac{6}{5}, 0.4\right)\).

3. **Calculate the area of the shaded region \( \triangle AGE\):**

   Using the formula for the area of a triangle given its vertices \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\):
   \[
   \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
   \]

   Substituting the coordinates of \(A(0, 1)\), \(G\left(\frac{6}{5}, 0.4\right)\), and \(E\left(\frac{2}{3}, 0\right)\):
   \[
   \text{Area} = \frac{1}{2} \left| 0(0.4 - 0) + \frac{6}{5}(0 - 1) + \frac{2}{3}(1 - 0.4) \right|
   \]
   \[
   = \frac{1}{2} \left| 0 + \frac{6}{5}(-1) + \frac{2}{3}(0.6) \right|
   \]
   \[
   = \frac{1}{2} \left| -\frac{6}{5} + \frac{2 \times 0.6}{3} \right|
   \]
   \[
   = \frac{1}{2} \left| -\frac{6}{5} + 0.4 \right|
   \]
   \[
   = \frac{1}{2} \left| -\frac{6}{5} + \frac{2}{5} \right|
   \]
   \[
   = \frac{1}{2} \left| -\frac{4}{5} \right|
   \]
   \[
   = \frac{1}{2} \times \frac{4}{5}
   \]
   \[
   = \frac{2}{5}
   \]

The area of the shaded region is \( \frac{2}{5} \) square units.

Would you like any further details or have any questions?

Here are five more questions for you to consider:

1. How do you calculate the intersection of two lines algebraically?
2. Can you find the area of a triangle given its vertices?
3. What are the different forms of a line equation?
4. How do you determine the coordinates of a point that divides a line segment in a given ratio?
5. What is the significance of the slope of a line?

**Tip:** When solving geometry problems, drawing a clear diagram and labeling all known points and lines can greatly simplify the solution process.

Learn how to calculate the area of a shaded region in a unit square involving geometric properties and coordinate geometry. This problem demonstrates the use of coordinate geometry to find intersections and calculate triangle areas. Suitable for high school students.

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