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Question

A man started walking from his house towards south. After walking 6 km, he turned to his left and walked 5 km. Then he walked further 3 km after turning left. He then turned to his left and continued his walk for 9 km. How far is he away from his house?

This question was previously asked in
UGC NET 2014 Paper 1 Question Paper (28-Dec-2014)
The correct answer is

5 km

The safe method is to fix a coordinate frame, work out which way each left turn faces him, and add up the displacement along the two axes separately.

Let the house be the origin, with east positive along the horizontal axis and north positive along the vertical axis. Note the rule for turns: facing south, your left is east; facing east, your left is north; facing north, your left is west.

He walks 6 km south, so he is at 6 km south of the house.

He turns left, now facing east, and walks 5 km. He is 5 km east and 6 km south.

He turns left again, now facing north, and walks 3 km. His southward displacement falls from 6 to 3, so he is 5 km east and 3 km south.

He turns left a third time, now facing west, and walks 9 km. His eastward displacement of 5 is cancelled and he carries 4 km beyond, so he ends 4 km west and 3 km south of the house.

These two displacements are at right angles, so the straight line distance is the hypotenuse of a right triangle with legs 4 and 3. That is the square root of 16 + 9, the square root of 25, which is 5 km - the familiar 3-4-5 triangle.

The most common error is treating each left turn as if he were still facing his original direction, or adding the four distances. Tracking the facing direction at every turn is what prevents both.

Hence he is 5 km away from his house.

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