Question

a man goes 10 metres towards north then 20 metre towards east and the distance and displacement​

Answer 1

The path that the man followed would be as attached above! (click on the above image to open it in fullscreen!)

Let his initial position be A.

Let the distance he walked northwards be AB (= 10 m) and that he walked eastwards be BC (= 20 m).

Distance walked by the man

= (AB + BC) m

= (10 + 20) m

= 30 m

Displacement of the man

= AC

Now, ∆ABC is a right angled triangle, right angled at B. So by applying Pythagoras' theorem;

AC

$= (\sqrt{ {ab}^{2} + {bc}^{2} }) \: m \\ = (\sqrt{ {10}^{2} + {20}^{2} } ) \: m \\ = ( \sqrt{100 + 400}) \: m \\ = (\sqrt{500}) \: m \\ = ( \sqrt{2 \times 2 \times 5 \times 5 \times 5}) \: m \\ = (\sqrt{ {2}^{2} \times {5}^{2} \times 5 } ) \: m \\ = (2 \times 5 \sqrt{5} ) \: m \\ = 10 \sqrt{5} \: m$

Hence, the distance walked by the man and his displacement are 30 m and 10√5 m respectively.