Q3 of 95 Page 5

A man goes 15 m due west and then 8 m due north. How far is he from the starting point?


Let AB = 15m and AC = 8m


In ∆CAB, using Pythagoras Theorm,


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


⇒ (AC)2 + (AB)2 = (BC)2


⇒ (8)2 + (15)2 = (BC)2


⇒ (BC)2 = 64 + 225


⇒ (BC)2 = 289


⇒ BC = ±17


⇒ BC = 17 [taking positive square root]


Hence, the man is 17m far from the starting point.


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