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Question

Difficulty: MediumDirection and Distance Test

A land surveyor begins at a base camp OO. She walks 12 m12\text{ m} due East to reach point AA. From point AA, she turns 90∘90^\circ clockwise and walks 16 m16\text{ m} to point BB. Facing South at point BB, she turns 135∘135^\circ anti-clockwise and walks 102 m10\sqrt{2}\text{ m} to point CC. Finally, she turns due West and walks 14 m14\text{ m} to stop at point DD. What is the shortest distance (in meters) between the base camp OO and point DD?

Answer: 10 meters

Answer

The shortest distance between the base camp OO and point DD is 10 meters10\text{ meters}.
Tracking relative coordinates gives point A(12,0)A(12,0), point B(12,−16)B(12,-16), point C(22,−6)C(22,-6), and final point D(8,−6)D(8,-6). The straight-line distance from the origin O(0,0)O(0,0) to D(8,−6)D(8,-6) is 82+(−6)2=10 meters\sqrt{8^2 + (-6)^2} = 10\text{ meters}.

Step-by-Step Solution

1
Establish coordinate system
Base camp O=(0,0)O = (0, 0)
Assigning the origin to the start point allows vector tracking.
2
Calculate displacement to point A and B
Point A=(12,0)A = (12, 0), Point B=(12,−16)B = (12, -16)
East increases x-coordinate by 12; South decreases y-coordinate by 16.
3
Decompose angular movement from B to C
Point C=(22,−6)C = (22, -6)
Turning 135∘135^\circ anti-clockwise from South points to North-East. Δx=102×12=10 m\Delta x = 10\sqrt{2} \times \frac{1}{\sqrt{2}} = 10\text{ m} East, Δy=102×12=10 m\Delta y = 10\sqrt{2} \times \frac{1}{\sqrt{2}} = 10\text{ m} North.
4
Determine final coordinates at point D
Point D=(8,−6)D = (8, -6)
Walking 14 m14\text{ m} West reduces the x-coordinate from 22 to 8.
5
Apply distance formula for straight-line displacement
Distance OD=10 mOD = 10\text{ m}
OD=(8−0)2+(−6−0)2=64+36=10 mOD = \sqrt{(8 - 0)^2 + (-6 - 0)^2} = \sqrt{64 + 36} = 10\text{ m}.

Key Concept

Vector displacement decomposition and 2D Cartesian plane distance calculation
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Topics

Direction and Distance Test

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