Question

A baby crawls 12 feet towards east and then 4 feet towards south. He then crawls 9 feet towards west. Ho far is he from his initial position?

Answer 1

12 feet towards east.

So, imagine a path ABC from the start.

AB = 12

4 feet towards south, which means BC = 4

9 feet towards west, which means CA = 9.

The distance from his initial position will be a straight line (the Hypotenuse.)

So,

12 - 9 = 3

BC = 4.

By Pythagoras theorem

$3^{2} + 4^{2} = 9 + 16 = 25 \\ x^{2} = 25 \\ x = \sqrt{25} \\ x = 5$
Hence, the distance from his initial position is 5 units South-East.

Hope this helps!! :)

Answer 2

Answer:  

The baby is finally at 5 feet away from his initial position.

Step-by-step explanation:

Given a baby crawls from a point towards east by 12 feet.

And then to the south by 4 feet.

And again towards west by 9 feet.

The baby crawled 12 feet in forward direction, later 4 feet in downwards direction. As east and west are opposite directions, the baby has crawled back 9 feet.

Therefore, the extra distance remaining along west direction is $12-9=3feet$

As shown in the figure, ABC forms a right angled triangle, where AB is 3feet, and BC is 4feet.

Therefore, applying Pythagorean theorem, $c^{2} =\sqrt{a^{2} +b^{2} }$, we get

$AC^{2} =\sqrt{3^{2} +4^{2} } =\sqrt{9 +16} =\sqrt{25}$

$AC=\sqrt{25}=5feet$    

Therefore, the baby is finally at 5 feet away from his initial position.

For more info:

https://brainly.in/question/5196248

https://brainly.in/question/1157534