Question

a person Walks 4 kilometres towards west then turns to his right to travel 9 kilometres he turns towards east and travels 12 km finally travel 3 km towards south how far is he from the initial point​

Answer 1

Answer:

Use Pythagoras to get displacement

$x = \sqrt{ {6}^{2} + {8}^{2} } = \sqrt{100} = 10$

So he is in North-East , 10 m .

Answer 2

Answer:

Person is 10 kilometers away from his initial point.

Step-by-step explanation:

Consider that the point A is the initial point from where person starts to walk.

Then he walked the given distances in the given directions and reached at point E as shown in the attached figure.

The distance between initial point A and final point E = AE

We can find AE by drawing the right angle triangle ΔAEF,

From the Pythagoras theorem:

$AE= \sqrt{(AF)^{2}+(EF)^{2}}$                                               ..............(1)

From the diagram AF = $CD - AB = 12 - 4 = 8Km$

And, EF = $BC-DE= 9-3=6Km$

Put the value of AF and EF in eq. (1):

$AE= \sqrt{(8)^{2}+(6)^{2}}$

$AE= \sqrt{64+36}$

$AE=\sqrt{100}$

$AE=10Km$

Therefore, the distance between the initial point and final point is equal to 10km. So, the person is 10Km far from the initial point.