Question

A man goes 3 km due north and then 4 km due east. How far is he away from his initial position?

Answer 1

Suppose the man starts from A and goes 3 km north and reaches B.
He then goes 4 km towards east and reaches C.

∴ AB = 3 km
BC = 4 km
We have to find AC.
By Pythagoras theorem:
⇒$A{C}^2=A{B}^2+B{C}^2$
$\Rightarrow A{C}^2=3^2+4^2$
$=9+16$
$\Rightarrow A{C}^2=25$
$\Rightarrow AC=\sqrt{5^2}$
$\therefore AC=5km$
Hence, he is 5 km far from the initial position.