maths
posted by faizel .
Midpoints of the sides AB and AC of a triangle ABC are ( 3 , 5 ) and ( 3 , 3 ) respectively,then the length of the side BC is

let A = (h,k). Then
B=(h+2(3h),k+2(5k))=(6h,10k)
C=(h+2(3h),k+2(3,k))=(6h,6k)
BC^2 = 12^2+16^2 = 340
BC=√340