Question

Starting from Anil's house, Peter first goes 50 m to south, then 75 m to west, then 62 m to
North and finally 40 m to east and reaches Salim's house. Then find the distance between
Anil's house and Salim's house.

Answer 1

BA=(75-40)m=35 m

BS=(62-50)m=12m

therefore the displacement is=AS

AS²=AB²+BS²

=>AS²=35²+12²

=>AS=√1369

=>AS=37 metres

Answer 2

The distance between Anil's house and Salim's house=37 m

Step-by-step explanation:

BC=50 m

CD=75 m

FD=62 m

AF=40 m

CD=BE=75 m

BC=ED=50 m

EF=AG=FD-ED=62-50=12 m

EG=AF=40 m

GB=EB-EG=75-40=35 m

In triangle AGB

$AB^2=AG^2+BG^2$

By using Pythagoras theorem

$(Hypotenuse)^2=(Base)^2+(Perpendicular\;side)^2$

Substitute the values

$AB^2=(12)^2+(35)^2=144+1225$

$AB^2=1369$

$AB=\sqrt{1369}=37 m$

Hence, the distance between Anil's house and Salim's house=37 m

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