Question

三角形ABC是一个直角三角形,EG垂直于AC,四边形BDEF是一个正方形,EG=3厘米,AB、BC和AC分别等于30厘米

Answer 1

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