Question

Segment AN is an altitude of right ΔABC with a right angle at A. If AB = 2 sqrt5 in and NC = 1 in, find BN, AN, AC.

Answer 1

according to the right triangle altitude theorem, AN²=CN*NB=1*NB=NB
ΔANB is a right triangle, so AN²=AB²-NB²=(2√5)²-NB²=20-NB²

NB=20-NB²
solve for NB: NB²+NB-20=0, (NB+5)(NB-4)=0, NB=4 or NB=-5 (ignore this negative root)

AN²=NB=4, so AN=2

AC²=AN²+CN², AC²=2²+1², AC=√5