Question

. Saurabh started from place A and went to place B at a distance of 56 m due
east. From there he went to place C at a distance of 105 m due north. What is
the minimum distance between the places A and C? ​

Answer 1

Hope it helps you. Good Morning

Answer 2

The minimum distance between the places A and C is  119m

GIven Saurabh started at point A went to B at a distance of 56m due east.

Now he turns left and goes north by 105m.

Now the shortest distance between A and C will be the hypotenuse of the right angles triangle ABC, right angled at B.

AC = $\sqrt{AB^{2} + BC^{2} }$ = $\sqrt{56^{2} + 105^{2} }$

AC = $\sqrt{3136 + 11025}$ = $\sqrt{14161}$ = 119m

This can also be visualized as the triangle law,

Where the third side is always smaller than sum of the other two sides.

• ie, AB < AC + BC

•   AC < AB + BC

•   BC < AC + AB

Therefore the shortest distance between the start and the end point is 119m.