求教数学大神 。。。。
设A的边长为X,F的边长为Y,如图,那么
S0=75-X
S12=75-Y
S13=112-X-S12=112-X-(75-Y)=37-X+Y
S7=X-S13=X-(37-X+Y)=2X-Y-37
S10=X-S0=X-(75-X)=2X-75
S1=S7+S10=(2X-Y-37)+(2X-75)=4X-Y-112
S9=S12-Y=75-Y-Y=75-2Y
S11=S12-S13=(75-Y)-(37-X+Y)=X-2Y+38
S2=S11+S9=(X-2Y+38)+(75-2Y)=X-4Y+113
因为S1+S2=S13
所以(4X-Y-112)+(X-4Y+113)=37-X+Y,即X-Y=6,X=Y+6
S3=S1+S10=(4X-Y-112)+(2X-75)=6X-Y-187
S4=S0-S10-S3=(75-X)-(2X-75)-(6X-Y-187)=337-9X+Y
S5=S4-S3=(337-9X+Y)-(6X-Y-187)=524-15X-2Y=524-15(Y+6)-2Y=434-13Y
S6=Y-S9=Y-(75-2Y)=3Y-75
S8=S2+S9-S6=(X-4Y-112)+(75-2Y)-(3Y-75)=263+X-9Y=263+(Y+6)-9Y=269-8Y
因为S5=S8,所以:434-13Y=269-8Y,即Y=33,F边长为33
X=Y+6=36,A边长为36
B边长=S13=37-X+Y=31
C边长=S2=X-4Y+113=20
D边长=S3=6X-Y-187=14
E边长=S9=75-2Y=9
A+x+F=75
A+B+E+F=112
B+C+F-E=75
D+C+E+F+A-x=112
高人! 能说的详细点吗。。。谢谢