Question

A man goes 10 m due east and then 24m due north how far is he from the starting point

Answer 1

let initial position be A
after Moving 10m to east be B
And from B to north for 24 m be C
Thus
AB = 10
BC = 24
AC = under root of AB^2 + BC^2
AC = 26 m

Answer 2

First the man goes 10 m in north 
Then he goes 24 m in east

So the starting point, last point and the and the point where the man changes his direction forms a right angled triangle

The distance between starting point and last point = root(10^2 + 24^)
= root(100+576)
= root(676)
= 26 cm

This is your answer 

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