Question:
A man goes 10 m due south and then 24 due west. How far is he from the starting point?
Solution:
Let the man starts from point D and goes 10 m due south and stops at E. He then goes 24 m due west at F.
In right $\triangle \mathrm{DEF}$, we have:
DE = 10 m, EF = 24 m
$D F^{2}=E F^{2}+D E^{2}$
$D F=\sqrt{10^{2}+24^{2}}$
$=\sqrt{100+576}$
$=\sqrt{676}$
$=26 \mathrm{~m}$
Hence, the man is 26 m away from the starting point.