Question

Shyam, a fertilizer salesman, sells directly to farmers. He visits two villages A and B. Shyam starts from A, and travels 50 meters to the East, then 50 meters North-East at exactly 45° to his earlier direction, and then another 50 meters East to reach village B. If the shortest distance between villages A and B is in the form of $a\sqrt{b+\sqrt{c}}$ meters, Find the value of a+b+c.

Accepted Answer

E. Net horizontal displacement $x = 50 + 50\cos 45^\circ + 50 = 100 + 25\sqrt{2}$. Net vertical displacement $y = 50\sin 45^\circ = 25\sqrt{2}$. Shortest distance $d = \sqrt{x^2 + y^2} = \sqrt{(100 + 25\sqrt{2})^2 + (25\sqrt{2})^2} = \sqrt{10000 + 5000\sqrt{2} + 1250 + 1250} = \sqrt{12500 + 5000\sqrt{2}} = 50\sqrt{5 + 2\sqrt{2}} = 50\sqrt{5 + \sqrt{8}}$. So $a = 50, b = 5, c = 8$. $a+b+c = 50 + 5 + 8 = 63$. Since 63 is not among options A-D, option E (None of the above) is correct.

Shyam Salesman Walking Directions

Explanation

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