Math Problem Statement

解图中的数学题

Solution

The problem in the image involves finding the area of the shaded region in a unit square ABCDABCD where BE=2ECBE = 2EC and FF is the midpoint of CDCD.

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

  1. Identify and label the points:

    • AA is at (0,1)(0, 1)
    • BB is at (0,0)(0, 0)
    • CC is at (1,0)(1, 0)
    • DD is at (1,1)(1, 1)
    • FF is the midpoint of CDCD, so FF is at (1,0.5)(1, 0.5)
    • BE=2ECBE = 2EC, which means EE divides BCBC in the ratio 2:1. Given that BB is at (0,0)(0, 0) and CC is at (1,0)(1, 0), EE is at (23,0)(\frac{2}{3}, 0)
  2. Find the coordinates of GG, the intersection point of AFAF and DEDE:

    To find GG, we need the equations of lines AFAF and DEDE:

    • Equation of AFAF: The line AFAF passes through A(0,1)A(0, 1) and F(1,0.5)F(1, 0.5). The slope mAFm_{AF} is (0.51)/(10)=0.5(0.5 - 1) / (1 - 0) = -0.5. Using the point-slope form, the equation is y=0.5x+1y = -0.5x + 1.

    • Equation of DEDE: The line DEDE passes through D(1,1)D(1, 1) and E(23,0)E(\frac{2}{3}, 0). The slope mDEm_{DE} is (01)/(231)=3(0 - 1) / (\frac{2}{3} - 1) = -3. Using the point-slope form, the equation is y=3x+4y = -3x + 4.

    Setting the equations equal to each other to find the intersection: 0.5x+1=3x+4-0.5x + 1 = -3x + 4 Solving for xx:

    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 AGE\triangle AGE:

    Using the formula for the area of a triangle given its vertices (x1,y1)(x_1, y_1), (x2,y2)(x_2, y_2), (x3,y3)(x_3, y_3): Area=12x1(y2y3)+x2(y3y1)+x3(y1y2)\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)A(0, 1), G(65,0.4)G\left(\frac{6}{5}, 0.4\right), and E(23,0)E\left(\frac{2}{3}, 0\right): Area=120(0.40)+65(01)+23(10.4)\text{Area} = \frac{1}{2} \left| 0(0.4 - 0) + \frac{6}{5}(0 - 1) + \frac{2}{3}(1 - 0.4) \right| =120+65(1)+23(0.6)= \frac{1}{2} \left| 0 + \frac{6}{5}(-1) + \frac{2}{3}(0.6) \right| =1265+2×0.63= \frac{1}{2} \left| -\frac{6}{5} + \frac{2 \times 0.6}{3} \right| =1265+0.4= \frac{1}{2} \left| -\frac{6}{5} + 0.4 \right| =1265+25= \frac{1}{2} \left| -\frac{6}{5} + \frac{2}{5} \right| =1245= \frac{1}{2} \left| -\frac{4}{5} \right| =12×45= \frac{1}{2} \times \frac{4}{5} =25= \frac{2}{5}

The area of the shaded region is 25\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.

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Math Problem Analysis

Mathematical Concepts

Geometry
Coordinate Geometry
Area Calculation

Formulas

Area of a triangle given vertices

Theorems

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Suitable Grade Level

High School