Question

A man goes 24m due east and then 10m due north.How far is he away from his initial position

Answer 1

Hello dear


Here is your answer


A man goes to east = 24 m
And to north = 10 m

Distance = (24m)²+(10m)²
= 576m²+100m² = 676m²
= root of 676 m² = 26 m ans. ✅✅


Hence,
He is 26 m far away from his initial position.

If it helps you mark me as a brainlist

Thanks
❤️❤️

Answer 2

Given,

Distance covered in the east direction = 24 m

Distance covered in the north direction = 10 m

To find,

Distance from the initial position.

Solution,

We can easily solve this problem by following the given steps.

Now,

Let's take his initial position to be A, point in the east direction to be B and point in the north direction to be C.

We have to find the distance, AC.

Using Pythagoras theorem,

AC² = AB² + BC²

AC² = (24)²+(10)²

AC² = 576 + 100

AC² = 676

AC = √676

AC = 26 m.

( The distance due east and north are given in metres. So, the metre is also to be used here.)

Hence, he is 26 m far away from his initial position.