Question

A man goes 12m due east and then 5m due south.How far is he from the starting point?

Answer 1

displacement is given by
$s = \sqrt{ {12m}^{2} + {5m}^{2} } = 13m$

Answer 2

Answer:

ABC is right angled triangle ,

By Pythagoras property of triangle

AC² = BC² + AB²

AC² = 12² + 5²

AC²= 144 + 25

AC² = 169

AC²= 13²

AC = 13 cm

 Shortest distance from starting point = 13cm