Question

A man drives his car 5 km in east wart direction he turned right and went for 3 km then he turned west and drive for 1 km . how far is he from the starting point

Answer 1

minimum distance = √25 = 5 km
maximum distance = 9 km

Answer 2

Answer:

The man is 5km far from the starting point.

Step-by-step explanation:

Step 1:

Let the man start driving from point A.

He first drives 5km in the eastward direction, i.e., towards the right, and reaches point B.

From point B, he drives 3km towards the right, i.e., southwards, and reaches point C.

At point C, he turns west, i.e., towards the left, drives 1km, and reaches point D.

Step 2:

From point D, draw a perpendicular DE to the line AB. The length of segment EB will be 1km.

So, the length of segment AE will be (5-1) = 4km

Join points A and D to form a right-angled triangle AED.

Step 3:

In ΔAED, AE = 4 and ED = BC = 3

Using Pythagoras theorem, we get

$AD^{2} = AE^{2} + ED^{2}$

$AD^{2} = 4^{2} + 3^{2}$

$AD^{2} = 16 + 9$

$AD^{2} = 25$

Taking positive square roots on both sides.

AD = 5km

Therefore, after driving in the given directions, the man is now 5km from the starting point.

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