Question

矩形ABCD中有一点P到ABC的距离为1.2.3求P到点D的距离

Answer 1

设P到AB,CD的距离为a,b

P到BC ,AD的距离为m,n

则a^2+m^2=BP^2=4

a^2+n^2=1

b^2+m^2=9

b^2+n^2=PD^2

m^2-n^2=3 m^2-n^2=9-PD^2

所以9-PD^2=3

PD^2=6

PD=根号6