Previously, a fifth-grade competition question was released: only the difference between the area of two squares is known, and the area of an irregular quadrilateral is found! This problem is not difficult! The method is not right, and it is useless to break the head, so you can only hand in a blank paper; if the method is correct, it is clear at a glance and the answer is palatable!
Question 465: As shown in Figure 1,
Figure 1
The area difference between the two squares ABCD and CEFG is 10, and the area of the quadrilateral ABFG in the shaded part is found.
Many students think: this problem lacks conditions and cannot be solved!
Difficulty: It is impossible to find the area of an irregular quadrilateral by finding the respective side lengths or areas of two squares! In fact, the side length and area of these two squares cannot be found! This is also the reason why students think that "this question lacks conditions"!
If you want to specify the side length or area of two squares, you need to attach at least one condition: the difference between the side lengths of the two squares or the sum of the side lengths!
Analysis 1: Graphic rotation!
(1) Rotate the quadrilateral BCEF 90° counterclockwise around point C until CB coincides with CD and CE coincides with CG, and the rotated points G and F are denoted as points G' and F' respectively, and F'D and F'G are connected. See Figure 2
Figure II
(2) Obviously, the square CGF 'G' is the same as the square CEFG, the quadrilateral BCEF is the same as the quadrilateral CGF'D, and the △BGF is the same as the △DG 'F'. F, G, F' are in a straight line, which can also be understood as F' falling on the extension line of FG.
(3) Extend FG and intersect with AD at point H. As shown in Figure 3
Figure III
(4) S square ABCD-S square CEFG = S square ABCD-S square CGG'F' = S rectangle ABGH+S rectangle DG'F'H = 10.
(5) The area of the rectangle is bisected by the diagonal, and S△ABG=1/2S rectangular ABGH, S△DF'G'=1/2S rectangular DG'F'H can be obtained. Therefore, S△ABG+S△DF'G'=5.
(6) Note that △BGF is the same as △DG'F', i.e., S△BGF=S△DG'F'. Therefore, S shadow = S△ABG + S△BFG = S△ABG + S△DF'G' = 5.
Analysis 2: Graphic folding!
Fold the square CEFG along the CG into the square ABCD, you can also get Figure 3, and the rest of the steps are the same as analysis 1!
Analysis 3: Partitioning Method!
Split a square identical to CEFG in the lower right corner of the square ABCD, which can also be obtained in Figure 3, and the rest of the steps are the same as in analysis 1.
Note: The three types of analysis are essentially the same, but the angle of looking at the problem is slightly different!
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