Question

A man walks 2 km towards north. then he turns to east and walks 10 km. after this he turns to north and walks 3 km. again he turns towards east and walks 2 km. how far is he from the starting point

Answer 1

he is 13 km away from the starting point .
see the image.

Answer 2

Answer:

13 km

Step-by-step explanation:

Initially man walks 2 km to north i.e.AB = 2 km

Then he turns to east and walk 10 km i.e. BC =10 km

Then again walks 3 km to north i.e.CD = 3 km

Then again he turns to east and walk 2 km i.e. DE =2 km

we are supposed to find  how far is he from the starting point i.e. AE

Refer the attached figure

In ΔABC

$Hypotenuse^2=Perpendicular^2+Base^2$

$AC^2=AB ^2+BC^2$

$AC^2=2^2+10^2$

$AC^2=4+100$

$AC=\sqrt{104}$

In ΔCDE

$Hypotenuse^2=Perpendicular^2+Base^2$

$CE^2=CD ^2+DE^2$

$CE^2=3^2+2^2$

$CE^2=9+4$

$CE=\sqrt{13}$

AE = AC+CE = $\sqrt{104}+\sqrt{13}=13.803$

Thus he is 13 km far from the starting point .