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

26 m

Let A be the initial position of the man. He goes 24m towards east i.e; B and the he move 10m towards North that is C.

Therefore, \( {AC }^{ 2} \) = \( { AB}^{2 } \) + \( { BC}^{2 } \)

= \( {AC }^{ 2} \) = \( { 24}^{2 } \) + \( { 10}^{2 } \)

= \( {AC }^{ 2} \) = 576 + 100

= \( {AC }^{ 2} \) =676

= AC = 26 m

Therefore, he is 26 m away from the initial point.