Question

A child is looking for his father. He went 90 meters in the east before turning to his right. He went 20 meters before turning to is right again to look for his father at his uncle's place 30 meters from this point. His father was not there. From there, he went 100 meters to his north before meeting his father in a street. How far did the son meet his father from starting point ?

Answer 1

ANSWER
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The boy met his father 100 m away from the starting point

It is because on moving to his right 30 m for the second time the child counteracted 90 m east and only 60 m east resultant is left.

On moving to his first right 20 m he moved towards South and counteracted the 100 m North which he moved at last and only 80 m North resultant is left.

So, The Hypotenuse I.e. sqrt (80^2 + 60^2) is the distance from starting point hwich is equal to 100 m.


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HOPE IT HELPS :):):):)

Answer 2

Clearly, the child moves from A to B 90 metres eastwards upto B, then turns right and moves 20 metre upto C, then turns right and moves upto 30 metre upto D. Finally he turns right and moves upto 100 metre upto E.

So AB = 90 metre, BF = CD = 30 metre,

So, AF = AB - BF = 60 metre

Also DE = 100 metre, DF = BC = 20 metre

So, EF = DE - DF = 80 metre

as we can see in image that triangle AFE is a right angled triangle and we are having two sides, need to calculate third one, so we can apply Pythagoras theorem here

\begin{aligned}

A = AE = \sqrt{AF^2 + EF^2} \\

= \sqrt{(60)^2+(80)^2} \\

= \sqrt{3600+6400} \\

= \sqrt{10000} = 100

\end{aligned}

So from starting point his father was 100 metre away.