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.