A person walks 1 km towards the East and then he turns to the South and walks 5 km. Again he turns to East and walks 2 km, after this, he turns to North and walks 9 km. Now, how far is he from his starting point?

Logic: The directions for the person is given and needs to draw a diagram and answer the question.

Let, the starting point be "O". From 'O' person will move to the east direction for 1 km which is the point 'A'.

Now, from Point 'A' he will move towards the south direction for 5 km which is point 'B'. Again he turns to East and walks 2 km

The person will be moving towards the north direction for 9 km. Let, this point be 'D' which is the final point.

Therefore, the final diagram is as follows:  

Now, we have to find the shortest distance between starting point 'D' and the final point 'A'.

So, we have to apply the Pythagoras theorem in triangle "ODE" which is a right-angle triangle.

\((Hypotenuse)^2 = (Perpendicular)^2 + (Base) ^2\)

\((OD )^2 = (OE)^2 + (DE)^2\)

\((OD)^2 = (3)^2 + (4) ^2\)

\((OD)^2 = 9 + 16\)

\((OD)^2 = 25 \)

\(OD = 5 \)

As the shortest distance obtained between O and D is 5km.

Hence, the correct answer is "5 km".