Question

A lady walks 8 km towards east, then 3 km towards south and then she walks 4 km towards west how far is she from his initial position and in which direction?

Answer 1

Hey friend

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Answer 2

Answer:

She is 5 km from his initial position and in SOUTH EAST direction.

Step-by-step explanation:

Refer the attached figure

A lady walks 8 km towards east i.e. AB = 8 km

Then She walks 3 km towards south i.e. BC = 3 km

Then She walks 4 km towards west i.e. CD =4 km

Now we are supposed to find how far is she from his initial position i.e.AD

ED = BC = 3 km

EB = DC = 4 km

AB - EB = AE

8 km - 4 km = AE

4 km = AE

In ΔAED

$Hypotenuse^2=Perpendicular^2+Base^2$

$AD^2=AE^2+ED^2$

$AD^2=4^2+3^2$

$AD^2=16+9$

$AD^2=25$

$AD=\sqrt{25}$

$AD=5$

Hence she is 5 km from his initial position and in SOUTH EAST direction.