# A man goes 10 m due south and then 24 m due west. How far is he from the starting point?  Let AB = 10m and AC = 24m

In ∆CAB, using Pythagoras Theorem,

(Perpendicular)2 + (Base)2 = (Hypotenuse)2

(AC)2 + (AB)2 = (BC)2

(24)2 + (10)2 = (BC)2

(BC)2 = 576 + 100

(BC)2 = 676

BC = ±26

BC = 26 [taking positive square root]

Hence, the man is 26m far from the starting point.

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