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Rahul facing North, walks 2.8 Km at 45° to his right. Then turns right and walks 8Km, then turns right and walks 8Km, then turns right and walks 5Km. From there he walks 2.8Km at 45° to his right and stops. What is the distance between his starting point and ending point?
asked in Quantitative Aptitude by (48 points) 2 4
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Getting $6.003 \approx 6 km$ what is the answer?

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I'm rotating the reference frame by 45° anti-clockwise.

 

We're splitting UT into vertical and horizontal distance.

By  45-45-90 triangle we get UV = UT = 2.8 / √2 = 2.8/1.4 = 2 km.

The vertical distance between start point and end point is = 8 - 2.8 - 2

                                                                                               = 3.2 km

And the horizontal distance between start point and end point is = 8 - 5 - 2

                                                                                                      = 1 km

Therefore, The distance between starting and ending point will be the hypotenuse of right angle triangle with perpendicular sides 3.2 and 1.

∴ Required Distance = √ ( (3.2)2 + (1))

                             = √ (10.24 + 1)

                             = √ (11.24)

                             = 3.35

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