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Question
A boy first goes 5 m due north and then 12 m due east. Find the distance between the initial and the final position of the boy.
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Solution
Given: Direction of north = 5 m i.e. AC Direction of east = 12 m i.e. AB
To find: BC
According to Pythagoras Theorem,
In right angled Δ ABC
(BC)2 = (AC)2 + (AB)2
(BC)2 = (5)2 + (12)2
(BC)2 = 25 + 144
(BC)2= 169
∴ BC = `sqrt169=sqrt(13xx13)` = 13 m
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