A man starts walking from point $ A $ and reaches point $ \mathrm{D} $. From $ \mathrm{A} $ he walks towards north for $ 800 \mathrm{~m} $ to reach point $ \mathrm{B} $. From $ \mathrm{B} $ he walks towards east for $ 500 \mathrm{~m} $ to reach point C. From $ \mathrm{C} $ he again walks towards north for $ 400 \mathrm{~m} $ to reach point $ D $. Find the direct distance between points $ \mathrm{A} $ and $ \mathrm{D} $.
Given:
A man starts walking from point \( A \) and reaches point \( \mathrm{D} \). From \( \mathrm{A} \) he walks towards north for \( 800 \mathrm{~m} \) to reach point \( \mathrm{B} \). From \( \mathrm{B} \) he walks towards east for \( 500 \mathrm{~m} \) to reach point C. From \( \mathrm{C} \) he again walks towards north for \( 400 \mathrm{~m} \) to reach point \( D \).
To do:
We have to find the direct distance between points \( \mathrm{A} \) and \( \mathrm{D} \).
Solution:
From the figure,
The direct distance between points \( \mathrm{A} \) and \( \mathrm{D} = \) Hypotenuse of triangle AED
By Pythagoras theorem,
$AD^2=AE^2+DE^2$
$AD^2=(500)^2+(800+400)^2$
$=250000+1440000$
$=1690000$
$=(1300)^2$
$AD=1300\ m$
The direct distance between points \( \mathrm{A} \) and \( \mathrm{D} \) is 1300 m.
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