AB=AD=AA1=1,∠BAD=∠BAA1=∠DAA1=60已知平行六面体ABCD-A1B1C1D1中,AB=AD=AA1=1,∠BAD=∠BAA1=∠DAA1=60º,则A1c

Question

AB=AD=AA1=1,∠BAD=∠BAA1=∠DAA1=60
已知平行六面体ABCD-A1B1C1D1中,AB=AD=AA1=1,
∠BAD=∠BAA1=∠DAA1=60º,则A1c

Answer 1

∠BAD=∠BAA1=∠DAA1=60º
AB=AD=AA1=1
ΔABD、ΔAA1B、ΔAA1D、ΔA1BD都是等边三角形
连接A1D、BD、A1B,A1D=BD=A1B
设BD中点为M,连接AM,A1M
则,A1M⊥BD,AM⊥BD
BD⊥AA1M平面
AM=CM=√3/2,A1M=√3/2,AA1=1,∠A1MC=180º-∠A1MA
cos∠A1MA=(3/4+3/4-1)/(2*3/4)=1/3
A1C^2=A1M^2+CM^2+2A1M*CM*cos∠A1MA
=3/4+3/4+2*(3/4)*(1/3)=3/2+1/2=2