A girl walks 200 metres towards East and then she walks 150m towards north find the distance of the girl from the starting point
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First the girl walks in East and after walking 200 metres in the same direction, then she turns towards North and walks a distance of 150 metres.
Hence, the path of the girl can be shown as a right angled triangle ABC as shown in the attachment.
Hence,
AB = 200 m
BC = 150 m
AC = ?
Since ∆ABC is a right - angled triangle,
We can apply Pythagoras Theorem here.
So,
AC ^ 2 = AB ^ 2 + BC ^ 2
AC ^ 2 = ( 200 ) ^ 2 + ( 150 ) ^ 2
AC ^ 2 = 40000 + 22500
AC ^ 2 = 62500
AC = √ ( 62500 )
AC = 250 m
Hence,
The distance of girl from the starting point = 250 m
Hence, the path of the girl can be shown as a right angled triangle ABC as shown in the attachment.
Hence,
AB = 200 m
BC = 150 m
AC = ?
Since ∆ABC is a right - angled triangle,
We can apply Pythagoras Theorem here.
So,
AC ^ 2 = AB ^ 2 + BC ^ 2
AC ^ 2 = ( 200 ) ^ 2 + ( 150 ) ^ 2
AC ^ 2 = 40000 + 22500
AC ^ 2 = 62500
AC = √ ( 62500 )
AC = 250 m
Hence,
The distance of girl from the starting point = 250 m
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