Question

she walks 60m west then 11m north. How far is she from her initial position.​

Answer 1

Answer:

61 m far from initial position .

Step-by-step explanation:

see the above image for explanation

thank you .

Answer 2

$\large\bold\orange{\underline{Answer:-}}$

$\tt\small\blue{She\:is\:61\:metres\:far\:from\:her\:initial \:point . }$

$\large\bold\green{\underline{Explaination:-}}$

Refer to the photo above. Let her initial point be x. Then she walks 60 m towards west, then she stops. Let her stopping point be y. Now, she walks towards north, then she stops. Now, let her stopping point be z. Then after all, her distance from the initial point is xz. Clearly, a direction forms a right angle to another direction. Then, angle y measures 90°.Then, Δxyz is a right angled triangle. So, to find the length of xz, we can apply Pythagoras theorem, as:-

$\sqrt{ {60}^{2} + {11}^{2} } = \sqrt{3600 + 121} = \sqrt{3721}$

$= > 61 \: metres$

So, she is 61 metres far from her initial point.

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