Question

A man goes 24 m due west and them 10 m due north. How far is he from the starting point?
(a) 34 m
(b) 17 m
(c) 26 m
(d) 28 m

Answer 1

(c) 26 m

Suppose, the man starts from point A and goes 24 m due west to point B. From here, he goes 10 m due north and stops at C.
In right triangle ABC, we have:
AB = 24 m, BC = 10 m
Applying Pythagoras theorem, we get:
$$\begin{gathered} A{C}^2=A{B}^2+B{C}^2={24}^2+{10}^2 \\ A{C}^2=576+100=676 \\ AC=\sqrt{676}=26 \end{gathered}$$