A girl walks 20 meters towards North. Then, turning to her left, she walks 50 meters. Then, turning to her right, she walks 40 metres. Again, she…

Let the initial position be at the origin with coordinates $(0, 0)$(0,0).

The girl walks 20 meters North:

Position: $(0, 20)$(0,20)

Direction: North

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She turns left (now facing West) and walks 50 meters:

Position: $(0 - 50, 20) = (-50, 20)$(0−50,20)=(−50,20)

Direction: West

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She turns right (now facing North) and walks 40 meters:

Position: $(-50, 20 + 40) = (-50, 60)$(−50,20+40)=(−50,60)

Direction: North

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She turns right again (now facing East) and walks 50 meters:

Position: $(-50 + 50, 60) = (0, 60)$(−50+50,60)=(0,60)

Direction: East

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The distance from her initial position $(0, 0)$(0,0) to final position $(0, 60)$(0,60) is:

$$d = \sqrt{(0-0)^2 + (60-0)^2}$$d=(0−0)2+(60−0)2​

$$d = \sqrt{0 + 3600}$$d=0+3600​

$$d = \sqrt{3600}$$d=3600​

$d = 60$d=60 meters

Therefore, she is 60 meters from her initial position.