Kishore walks 10 ,km towards north. Then he w | vedclass

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Question

(C) Let the starting point be $A$. Kishore walks $10 \,km$ north to point $B$. Then he walks $6 \,km$ south to point $C$. So,the distance $AC = AB - B

Vedclass · Accepted Answer

$5 \,km$ north-east

Vedclass · Answer

(C) Let the starting point be $A$. Kishore walks $10 \,km$ north to point $B$. Then he walks $6 \,km$ south to point $C$. So,the distance $AC = AB - BC = 10 \,km - 6 \,km = 4 \,km$ (towards north). Then he walks $3 \,km$ towards east to point $D$. Now,we have a right-angled triangle $ACD$ where $AC = 4 \,km$ and $CD = 3 \,km$. The distance from the starting point $A$ is the hypotenuse $AD$. $AD = \sqrt{AC^2 + CD^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \,km$. Since point $D$ is to the north and east of point $A$,the direction is north-east. Thus,Kishore is $5 \,km$ north-east from his starting point.

Kishore walks $10 \,km$ towards north. Then he walks $6 \,km$ towards south. Then he walks $3 \,km$ towards east. How far and in which direction is he with reference to his starting point?

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