GATE Civil 2015 Set 2 | GA | Question: 4

Question

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?

• A. $2\sqrt{2}$
• B. $2$
• C. $\sqrt{2}$
• D. $\dfrac{1}{\sqrt{2}}$

Answer 1

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 

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

$\qquad = 4+4 = 8$

$\implies AE=2\sqrt 2.$

Answer 2

$\text{In } \triangle ASD$
$\cos 60^\circ = \frac{SD}{6\text{m}} \implies \boxed{SD = 3\text{m}}$

$\text{In } \triangle ABD$
$\sin 30^\circ = \frac{BD}{6\text{m}} \implies \boxed{BD = 3\text{m}}$

$FE = 2\text{m}$ ($\because FC = 6\text{m}$ and $EC = 4\text{m}$)

$\text{and }SF = BC = 2\text{m}$ ($\because SF \parallel BC$)

$\text{Using Pythagoras theorem in $\triangle SEF :$ }$
$SE = \sqrt{2^2 + 2^2} = 2\sqrt{2}$

$ \boxed{SE = 2\sqrt{2}\text{m}}$