Question

等了3天还没解出得数学几何题在正三棱锥A-BCD中,底边边长为a,侧棱长2a,E,F分别为AC,AD上得动点,求截面三角

Answer 1

BC=a ,B'D=a ,CD=a

∠ACD=Θ

cosΘ=1/2*CD/AC=1/4

∠CBB'=180-2Θ

过C,D作BB'垂线,垂足是P,Q

BP=BCcos(180-2Θ)

=a*sin2Θ

=a*2sinΘcosΘ

=√15/8*a

BB'=BP+B'Q+PQ

=15/4*a+a