Question

A man walks 2 km to east and then he turns to 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?
a)4.66Km
b)3.66km
c)5.66km
d)5.02km​

Answer 1

5.66

please mark me as brilliant

Answer 2

The distance between starting and ending points.is 5.66km(Option C).

Given,

A man walks 2 km to the east, turns to the south, and walks 5 km.

Then turns to the east, follows by 2 km walks then turns to the north, and walks 9 km.

To Find,

The distance between starting and ending point.

Solution,

We can solve this numerical question using the following method.

To solve this problem concepts we are using here are as follows,

• The Pythagoras theorem claims that the square of the hypotenuse of a right-angled triangle is equal to the sum of the squares of the other two sides.

• The hypotenuse is the shortest path between the two locations on either end of a right triangle, despite being the longest side.

Let's draw the image and denote the points.

We get, AB = 2kms, BC=5kms, CD=2kms, DE=9kms.

The final two points are A and E, so the distance be AE.

Now if we try to draw a perpendicular from B on the line, DE we get, BF. The distance of BF surely is 2km.

From the picture, it is clear that, AF=AB+BF

So, AF=2+2=4kms.

Also then, BC=FD=5kms.

This means EF=ED-FD.

⇒EF=9-5=4kms.

Now we get a right angle triangle ΔAFE, where AE is the hypotenuse, also AF=4, FE=4

∴$AE=\sqrt{(AF^2+EF^2)}$

⇒AE=$\sqrt{4^2+4^2}$.

⇒AE= $\sqrt{32}$.

⇒AE=5.66(Approx).

Hence, The distance between starting and ending points.is 5.66km(Option C).

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