Question

点M为矩形ABCD的边BC上的一点,DN垂直于AM于N,AB=3,BC=7,AM=5,则DN=?

Answer 1

用欧古定理求出BM,AM^2=AB^2+BM^2,5^2=3^2+BM^2,BM=根号16,取正,BM=4,BM+MC=BC=7,MC=7-4=3.

连结MD,Sabcd=s直角三角形abm+s三角形amd+s直角三角形mcd

3*7 =1/2*3*4+1/2*5*ND+1/2*3*3

21=6+5ND/2+9/2

5ND=21

ND=4.2