Kabita walks from her office 2 km to the east, then 8 km towards her right, then 2 km towards the east and then 4 km Kabita walks from her office 2 km to the east, then 8 km towards her right, then 2 km towards the east and then 4 km towards the south. Finally she turns left and walks 1 km to reach her apartment. The shortest aerial distance between Kabita’s office and her apartment is
Question:
Kabita walks from her office 2 km to the east, then 8 km towards her right, then 2 km towards the east and then 4 km towards the south. Finally she turns left and walks 1 km to reach her apartment. The shortest aerial distance between Kabita’s office and her apartment is:
(A) 25 km
(B) 13 km
(C) 12 km
(D) 17 km
Solution:
Given,
- Kabita walks 2 km east.
- Then, she turns right (south) and walks 8 km.
- Next, she walks 2 km east again.
- Then, she turns south and walks 4 km.
- Finally, she turns left (east) and walks 1 km.
Calculate the Net Displacement:
- Total eastward distance: km
- Total southward distance: km
Apply the Pythagorean Theorem:
Net displacement from the starting point is the hypotenuse of a right triangle with legs of 5 km (eastward) and 12 km (southward):
√(5²+12²)
√(25+ 144)
√169
13km
So, the shortest aerial distance between Kabita's office and her apartment is:13 km
Answer: (B) 13 km
Importance for ADRE 2.0
- Importance: Tests understanding of displacement and Pythagorean theorem.
- Enhances spatial reasoning and problem-solving skills.
- Crucial for competitive exam preparation, involving real-world scenarios.
- Previously asked in ADRE 1.0 SLRC-2022 for Grade-III posts of Assam Govt.
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