Question

如图所示,梯形ABCD中AB//CD,∠ADC=60°,∠BCD=30°,以AD,AB,BC向梯形外作正方形，它们面积分别为

S1，S2，S3，若DC=2AB，S2=27，求根号S1，根号S3.

Answer 1

S2=AB^2=27  AB=3√3

BE=CE*tan30=√3CE/3  AF=DF*tan60=√3DF

√3CE/3 =√3DF  CE=3DF

CD=2AB  EF=AB  CE+DF=AB

4DF=AB  DF=AB/4=3√3/4   CE=9√3/4

AD=DF/cos60=3√3/4*2=3√3/2

BD=CE/cos30=9√3/4*2/√3=9/2

S1=AD^2  √S1=AD=3√3/2

S3=BC^2   √S2=BC=9/2

Answer 2

图在哪里？