如图,直角三角形ABC内有一个正方形BDEF.已知AB=3厘米,BC=4厘米,AC=5厘米,EG垂直于AC,EG=0.3厘米,求

设正方形边长为x,连接AE,CE
则AF=3-x CD=4-x
S△ABC=S△AFE+S正方形BDEF+S△DCE+S△AEC
1/2AB×AC=1/2AF×EF+BF×BF+1/2CD×DE+1/2AC×EG
1/2×4×5=1/2(3-x)x+x²+1/2(4-x)x+1/2×5×0.3
10=3/2x-1/2x²+x²+2x-1/2x²+0.75
7/2x=10-0.75
7/2x=9.25
x=18.5/7
x=37/14
∴S正方形BDEF
=x²
=(37/14)²