Question

四棱锥P-ABCD的底面为矩形,PA垂直于平面ABCD,PB=根号5,PC=根号17,PD=根号13,求P到BD距离

Answer 1

PB^2=PA^2+AB^2=5

PD^2=PA^2+AD^2=13

PC^2=PA^2+AC^2=PA^2+AB^2+AD^2=17

得：PB^2+PD^2-PC^2=PA^2=1,PA=1

AB=2,AD=2√3,AC=4

作AE垂直BD于E,则AE=AB*AD/BD=AB*AD/AC=√3

P到BD距离PE=√(PA^2+AE^)=√(1+3)=2