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Mr. Vivek walks $6$ meters North-east, then turns and walks $6$ meters South-east, both at 60 degrees to east. He further moves $2$ meters South and $4$ meters West. What is the straight distance in meters between the point he started from and the point he finally reached?

- $2\sqrt 2$
- $2$
- $\sqrt 2$
- $1/\sqrt2$

11 votes

Best answer

Option A $2\sqrt 2$

$AB = 6m , \ BC = 6m , \ CD = 2m, \ ED = 4m$

$\angle BAC = 60^{\circ}, \angle BCA = 60^{\circ}$

$so, \angle ABC = (180^{\circ}-60^{\circ}-60^{\circ}) = 60^{\circ}$

and $AC = 6m (\text{Equilateral triangle})$

$A$ is the starting point. Vivek walks $6$ meters North-East to point $B,$ then turns and walks $6$ meters South-East to point $C.$ He further moves $2$ meters South to the point $D$, $4$ meters West to point $E$, and $E$ is the endpoint.

As shown in figure, $AC=6m$ and $ED= FC = 4m.$

$\implies AF=AC- FC = 6-4=2m$

and $EF= CD = 2m$ and $\angle AFE = 90^{\circ}$

By Pythagoras theorem

$AE^2 = EF^2 + AF^2$

$\qquad = 4+4 = 8$

$\implies AE=2\sqrt 2.$