Reena starts walking towards west and stops after 15 Km and turns left and…
Reena starts walking towards west and stops after 15 Km and turns left and walks another 10 Km and again turns left and walk 10 Km and stops. What is the minimum distance Reena has to walk to reach starting point ?
- D.
25 km
Attempted by 1 students.
Show answer & explanation
In direction-and-distance problems, assign each cardinal direction a perpendicular axis - East-West as one axis, North-South as the other - and track displacement as vector addition along these axes. A left turn rotates the direction of travel 90 degrees counter-clockwise, cycling West to South to East to North to West. Once the net East-West and net North-South displacements are known, the straight-line (minimum) distance back to the start is the hypotenuse of a right triangle formed by those two net legs, given by the Pythagorean theorem.
Applying this to Reena's walk:
Start at the origin (0, 0).
Walk 15 km West: position becomes (-15, 0).
Turn left (facing West, a left turn points South) and walk 10 km: position becomes (-15, -10).
Turn left again (facing South, a left turn points East) and walk 10 km: position becomes (-5, -10).
Net displacement from the start: 5 km West and 10 km South.
Apply the Pythagorean theorem to these two net legs to get the straight-line distance back to the start. Distance = √(52 + 102) = √(25 + 100) = √125 = 5√5 km.
Cross-check: the resulting distance is between the longer individual leg (10 km) and the total ground covered (15 + 10 + 10 = 35 km), exactly as the triangle inequality requires for a straight-line distance between two points connected by a multi-segment path - confirming the figure is consistent.