UPSC Prelims 2014·CSAT·Quantitative Aptitude·Geometry and Mensuration

Location of B is north of A and location of C is east of A. The distances AB and AC are 5 km and 12 km respectively. The shortest distance (in km) between the locations B and C is

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Last updated 23 May 2026, 3:31 pm IST
  1. A60
  2. B13Correct
  3. C17
  4. D7

Explanation

Explanation: 1. Visualize the locations: Location A is the reference point. B is directly north of A, and C is directly east of A. This arrangement forms a right-angled triangle with the right angle at A. 2. Identify the sides: The distance AB (5 km) is one leg of the triangle, and the distance AC (12 km) is the other leg. The shortest distance between B and C is the hypotenuse of this right-angled triangle. 3. Apply Pythagorean theorem: For a right-angled triangle, a^2 + b^2 = c^2, where 'c' is the hypotenuse. * BC^2 = AB^2 + AC^2 * BC^2 = 5^2 + 12^2 * BC^2 = 25 + 144 * BC^2 = 169 * BC = sqrt(169) * BC = 13 km Therefore, the shortest distance between B and C is 13 km. Option Analysis: A) 60: Incorrect. This would be the product of the distances (5 * 12), not the shortest distance. B) 13: Correct. This is calculated using the Pythagorean theorem. C) 17: Incorrect. This would be the sum of the distances (5 + 12), which is the path distance, not the shortest straight-line distance. D) 7: Incorrect. This would be the difference of the distances (12 - 5), which is not relevant here. The final answer is B.
Quantitative Aptitude: Location of B is north of A and location of C is east of A. The distances AB and AC are 5 km and 12 km respectively. The
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