如图,在正方形ABCD中,点E、F、G、H均在其内部,且DE=EF=FG=GH=HB=2,∠E=∠F=∠G=∠H=60°.如图,在正方形ABCD中,点E、F、G、H均在其内部,且DE=EF=FG=GH=HB=2,∠E=∠F=∠G=∠H=60°,则正方形ABCD的边长AB=------
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/04 23:47:35
![如图,在正方形ABCD中,点E、F、G、H均在其内部,且DE=EF=FG=GH=HB=2,∠E=∠F=∠G=∠H=60°.如图,在正方形ABCD中,点E、F、G、H均在其内部,且DE=EF=FG=GH=HB=2,∠E=∠F=∠G=∠H=60°,则正方形ABCD的边长AB=------](/uploads/image/z/11518376-32-6.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E6%AD%A3%E6%96%B9%E5%BD%A2ABCD%E4%B8%AD%2C%E7%82%B9E%E3%80%81F%E3%80%81G%E3%80%81H%E5%9D%87%E5%9C%A8%E5%85%B6%E5%86%85%E9%83%A8%2C%E4%B8%94DE%3DEF%3DFG%3DGH%3DHB%3D2%2C%E2%88%A0E%3D%E2%88%A0F%3D%E2%88%A0G%3D%E2%88%A0H%3D60%C2%B0.%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E6%AD%A3%E6%96%B9%E5%BD%A2ABCD%E4%B8%AD%2C%E7%82%B9E%E3%80%81F%E3%80%81G%E3%80%81H%E5%9D%87%E5%9C%A8%E5%85%B6%E5%86%85%E9%83%A8%2C%E4%B8%94DE%3DEF%3DFG%3DGH%3DHB%3D2%2C%E2%88%A0E%3D%E2%88%A0F%3D%E2%88%A0G%3D%E2%88%A0H%3D60%C2%B0%2C%E5%88%99%E6%AD%A3%E6%96%B9%E5%BD%A2ABCD%E7%9A%84%E8%BE%B9%E9%95%BFAB%3D------)
如图,在正方形ABCD中,点E、F、G、H均在其内部,且DE=EF=FG=GH=HB=2,∠E=∠F=∠G=∠H=60°.如图,在正方形ABCD中,点E、F、G、H均在其内部,且DE=EF=FG=GH=HB=2,∠E=∠F=∠G=∠H=60°,则正方形ABCD的边长AB=------
如图,在正方形ABCD中,点E、F、G、H均在其内部,且DE=EF=FG=GH=HB=2,∠E=∠F=∠G=∠H=60°.
如图,在正方形ABCD中,点E、F、G、H均在其内部,且DE=EF=FG=GH=HB=2,∠E=∠F=∠G=∠H=60°,则正方形ABCD的边长AB=------
如图,在正方形ABCD中,点E、F、G、H均在其内部,且DE=EF=FG=GH=HB=2,∠E=∠F=∠G=∠H=60°.如图,在正方形ABCD中,点E、F、G、H均在其内部,且DE=EF=FG=GH=HB=2,∠E=∠F=∠G=∠H=60°,则正方形ABCD的边长AB=------
设角ABH=T
AD=BH*sinT+HGsin(60-T)+GFsinT+FEsin(60-T)+DEsinT
=2(3sinT+2sin(60-T))
AB=BH*cosT+GFcosT+DEcosT-GHcos(60-T)-EFcos(60-T)
=2(3cosT-2cos(60-T))=AD
3cosT-2cos(60-T)=3sinT+2sin(60-T)
sin(60-T)+cos(60-T) = 3/2 (cosT-sinT)
sqrt(2) sin(105-T) = 3/2 sqrt(2) cos(T-45)
sin(105-T) = 3/2 cos(T-45)
T = 4.105
AB = 3.74
连结DF,FH,EG,GB,BD交FG于O,则三角形DEF,EFG,FGH,GHB都是等边三角形,四边形BEDH是平行四边形,所以点O是FG的中点,在三角形DOF中,OF=1,DF=2,角DFO=120度,所以OD=根号7,BD=2倍根号7,AB=根号14.