Question

A man walks $5m$due south and then $12m$due east. Find his distance from the starting point.

Answer 1

Step 1: Draw the figure from the given expression

Given, $5msouthand12meast$

Let the starting point is $A$

He goes $5msouth$ then $AB=5m$

Now, he goes $12meast$ then $BC=12m$

So there is a right-angled triangle formed.

Step 2:We have to find $AC$

By the Pythagoras theorem, in $△ABC$ we have

$$\begin{gathered} A{C}^2=A{B}^2+B{C}^2 \\ =5^2+{12}^2 \\ =25+144=169 \\ \Rightarrow AC=\sqrt{169}=13 \end{gathered}$$

So, $AC=13$

Hence, He is$13m$ from the starting point.