Question

三棱锥P-ABC中,D为AC中点,PA=PB=PC=根号5,AC=2倍根号2,AB=根号2,BC=根号6,

Answer 1

1作AB中点E,

AC=2√2,AB=√2,BC=√6,AC^2=AB^2+BC^2,D是斜边AC中点,AD=BD=CD,D是ABACBC垂直平分线的交点,DE垂直AB,PA=PB.PE垂直AB,AB垂直平面PDE,PD垂直于AB,PA=PC,PD垂直于AC,PD垂直于平面ABC

2

DE=BC/2=√6/2,PD^2=PA^2-PAD^2=5-2=3,PD=√3

PD/DE=√3/(√6/2)=√2

tg二面角P-AB-C=√2

3

PE^2=PB^2-BE^2=5-1/2=9/2 PE=3/2

Sbpe=BE*PE/2=PB*H/2

H=BE*PE/PB=3√(2*/5) /4=3√10/20

E到PBC距离=3√10/20