A man goes 24 m due east and then 10 m due north. How far is he away from his initial position?
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A man goes 24 m due east and then 10 m due north. How far is he away from his initial position?
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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.
Given:
Find:
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
[tex] \rm \implies{OB }^{2} = ({OA }^{2} + {AB }^{2} )[/tex]
Where,
Now,
[tex] \rm \implies{OB }^{2} = \{{(24)}^{2} + {(10) }^{2} ) \} {m}^{2} [/tex]
[tex] \rm \implies{OB }^{2} = (576 + 100) {m}^{2} [/tex]
[tex] \rm \implies{OB }^{2} = 676 {m}^{2} [/tex]
[tex] \rm \implies OB = \sqrt{676}m = 26m[/tex]
[tex] \rm \to So, OB = 26m[/tex]
Hence, The Man Is At Distance Of 26m From His Intial Position.