Question 102193
let AR=a
a^2+4^2=8^2 pythagoras a^2+(BR)^2=(AB)^2
a^2+16=64
a^2=48
a=sqrt(16*3)=4*sqrt(3)
so ER=4sqrt(3)/2=2sqrt(3)
Let c=BE
4^2+(2sqrt(3))^2=c^2 pythagorus (BR)^2+(ER)^2=c^2
16+12=c^2
c^2=28
c=sqrt(7*4)=2sqrt(7)
Ed