Question

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

Answer 1

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Answer :

26m

Solution :

Given,

Man goes due east for 24 m.

and due north 10 m.

This forms a right angled triangle whose sides are 24 m and 10 m.

The distance between his initial and final positions is the length of hypotenuse of the triangle.

By Pythagoras theorem,

hypotenuse square = sum of squares of other 2 sides

d² = 24² + 10²

d² = 676

d = √676

d = 26

Therefore,

Distance between initial and final points is 26 m.

Answer 2

Given:

• A Man Goes 24m due East
• And 10m due North

Find:

• How Far He Is From Initial Point

Solution:

Here,

Let, O Be The Initial Position Of The Man

Now, Let Him Cover OA = 24m due East and Then AB = 10m due North.

Finally He Reaches The Point B. Join OB.

Now, In Right Triangle OAB, we have

$\rm \implies{OB }^{2} = ({OA }^{2} + {AB }^{2} )$

Where,

• OA = 24m
• AB = 10m

Now,

$\rm \implies{OB }^{2} = \{{(24)}^{2} + {(10) }^{2} ) \} {m}^{2}$

$\rm \implies{OB }^{2} = (576 + 100) {m}^{2}$

$\rm \implies{OB }^{2} = 676 {m}^{2}$

$\rm \implies OB = \sqrt{676}m = 26m$

$\rm \to So, OB = 26m$

Hence, The Man Is At Distance Of 26m From His Intial Position.