A man starts from his home and walks 10 , km | vedclass

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(C) Let the starting point be $A$. $1$. The man walks $10 \, km$ South to reach point $B$. $2$. He turns right (West) and walks $6 \, km$ to reach poi

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(C) Let the starting point be $A$. $1$. The man walks $10 \, km$ South to reach point $B$. $2$. He turns right (West) and walks $6 \, km$ to reach point $C$. $3$. He turns right (North) and walks $10 \, km$ to reach point $D$. $4$. Finally,he turns right (East) and walks $5 \, km$ to reach point $E$. $5$. The distance from the starting point $A$ is $AE = AB - CD$ (if we consider the horizontal displacement) or more simply,looking at the diagram: Horizontal distance from $A$ to $E$ is $6 \, km - 5 \, km = 1 \, km$. Thus,the man is $1 \, km$ away from his starting point.

$A$ man starts from his home and walks $10 \, km$ towards South. Then he turns right and walks $6 \, km$,again he turns right and goes $10 \, km$. Finally,he turns right and walks $5 \, km$. At what distance is he from his starting point? (in $km$)

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