Question

OP
Choose the correct option.
A child walks 6 m towards west and then 20 m towards south. He then turns east
and walks another 12 m. Then he walks 12 m towards north. How far is he from his
initial position?

Answer 1

Answer:

10m

Step-by-step explanation:

We can represent the position of the child with the Cartesian plane. Let the initial position of the child be at the origin

ATQ,

$x = 0 \: \: \: \: y = 0 \\ x = - 6 \: \: \: \: y = 0 \\ x = - 6 \: \: \: \: y = - 20 \\ x = 6 \: \: \: \: y = - 8$

Ultimately, the child ends up at the position (6,-8)

Now using the distance formula

$\sqrt{ {(x1 - x2)}^{2} + {(y1 - y2)}^{2} }$

Where,

x1 = 0, y1 = 0

x2 = 6, y2 = -8

$\sqrt{ {(0 - 6)}^{2} + {(0 - ( - 8))}^{2} } \\ \sqrt{ {6}^{2} + {8}^{2} } \\ \sqrt{36 + 64} \\ \sqrt{100} \\ 10$

Hence the child is 10m away from its initial position