Question

A man goes 15 metres due west and then 8 metres due north. How far is he from starting point?

Answer 1

First the person moves from A to B and then B to C.

His distance from the starting point is AC.

AC can be calculated using Pythagoras Theorem.

Applying Pythagoras theorem

$(AB{)}^2+(BC{)}^2=(CA{)}^2$

$(15{)}^2+(8{)}^2=(x{)}^2$

$225+64=(x{)}^2$

$x^2=289$

$x=\sqrt{289}=17m$

Hence the man is $17m$ from the starting point.