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Math problems in junior high school and primary school. The areas of square ABCD and EFGH are 9 square centimeters and 64 square centimeters respectively.
Solution:

The side length of square ABCD is √ 9 = 3cm,

The side length of the square EFGH = √ 64 = 8cm,

If AB and GF intersect m, the quadrilateral BCFM is rectangular,

BM=CF,MF = BC = 3cm,

Let BM=CF=X, then AM=AB+BM=3+X, GM=FG+MF=8+3= 1 1,

Shadow area = triangular AMG area-triangular ABC area-rectangular BCFM area-triangular CFG area

7.5 =(3+X)× 1 1÷2-3×3÷2-3×X-8×X÷2

7.5= 16.5+5.5X-4.5-3X-4X

1.5X=4.5

X=3

The area of the triangle BCF =BC×CF÷2=3×3÷2=4.5 (square centimeter).