(2) as shown in fig. 3, if O 1A 1 intersects with CB at point m and OA intersects with C 1B 1 at point n, the overlapping area of rectangle OA 1B 1 and rectangle OABC is the area of quadrilateral DNEM.
According to the meaning of the question, dm∨ne, dnem and ∴ quadrilateral DNEM are parallelograms.
According to the axial symmetry, ∠ med = ∠ ned.
And ∠ MDE = ∠ Ned, ∴ Med = ∠ MDE, ∴ MD = Me, ∴ parallelogram DNEM is a diamond.
The intersection d is DH⊥OA, and the vertical foot is H.
It is easy to know from the title that tan∠DEN= =, DH = 1, He = 2.
Let the side length of rhombic DNEM be a,
So in Rt△DHM, we know from Pythagorean theorem: Ⅷ.
∴S quadrilateral dnem = ne? DH= 1.25
∴ The overlapping area of rectangle OA1b1and rectangle OABC remains unchanged, and the area is always 1.25.
Similarly, you can also take a special value to do this problem, because the overlapping part is always a diamond, and when b= 1, you can also get the Pythagorean theorem equation.