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20 16 Putuo Senior Three Mathematics Module 2
Solution: solution: from the meaning of the question, point c and point h,

Point b and point g are symmetrical about the straight line EF,

∴CF=HF,BE=GE.

Let BE=GE=x, then AE = 4-X.

∵ quadrilateral ABCD is a square,

∴∠A=90。

∴AE2+AG2=EG2.

∵B falls at the midpoint g of the edge AD,

∴AG=2,

∴(4-x)2+22=x2.

Solution x = 2.5.

∴BE=2.5.

∵ quadrilateral ABCD is a square,

∴AB∥CD,∠B=90。

Points e and f are on the sides of AB and CD, respectively.

∴ Quadrilateral BCFE is a right-angled trapezoid.

BE = GE = 2.5,AB=4,

∴AE= 1.5.

∴sin∠ 1=35,tan∠ 1=34.

∵∠ 1+∠2=90 ,∠2+∠3=90 ,

∴∠3=∠ 1.

∴sin∠3=sin∠ 1=35,

In Rt△DGP, ∫∠d = 90,

DG=2,sin∠3=DGGP=35,

∴PG= 103,

∴PH=GH-GP=23,

∵∠4=∠3,

∴tan∠4=tan∠3=tan∠ 1=34,

In Rt△HPF, ∫∠h =∠c = 90,

∴FC=HF= 12.

∴S quadrilateral bcfe =12 (fc+be) × BC =12 × (12+2.5 )× 4 = 6.

So the answer is: 6.