A man walks 6 , km to the east and then turns | vedclass

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(C) Let the starting point be $A$. The man walks $6 \, km$ east to point $B$,then $5 \, km$ south to point $C$,then $6 \, km$ east to point $D$,and fi

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(C) Let the starting point be $A$. The man walks $6 \, km$ east to point $B$,then $5 \, km$ south to point $C$,then $6 \, km$ east to point $D$,and finally $10 \, km$ north to point $E$.To find the distance $AE$,we construct a right-angled triangle. Extend $AB$ to a point $O$ such that $EO$ is perpendicular to $AO$.Here,$AO = AB + BO = 6 \, km + 6 \, km = 12 \, km$ (since $BO = CD = 6 \, km$).The vertical distance $OE = ED - OD = 10 \, km - 5 \, km = 5 \, km$.Using the Pythagorean theorem in $	riangle AOE$:$AE^2 = AO^2 + OE^2$$AE^2 = 12^2 + 5^2$$AE^2 = 144 + 25 = 169$$AE = \sqrt{169} = 13 \, km$.Thus,the man is $13 \, km$ away from his starting point.

$A$ man walks $6 \, km$ to the east and then turns to the south and walks $5 \, km$. Again he turns to the east and walks $6 \, km$. Next,he turns towards north and walks $10 \, km$. How far is he now from his starting point (in $, km$)?

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