A man's house faces south. Starting from the front door, he walks straight for…
2023
A man's house faces south. Starting from the front door, he walks straight for 20 metres. He then turns left and walks 50 metres. Next, he turns right, walks 80 metres, and stops. Find the distance between his stopping point and the front door of his house.
Answer: B. \(50\sqrt{5}\,\text{m}\) — CONCEPTRepresent perpendicular movements as horizontal and vertical displacements. Movements in the same direction add, while opposite movements subtract. The…
- A.
\(25\sqrt{5}\,\text{m}\)
- B.
\(50\sqrt{5}\,\text{m}\)
- C.
50 m
- D.
100 m
Attempted by 1 students.
Show answer & explanation
Correct answer: B
CONCEPT
Represent perpendicular movements as horizontal and vertical displacements. Movements in the same direction add, while opposite movements subtract.
The straight-line distance from the start to the finish is the hypotenuse of a right triangle: d2 = x2 + y2.
APPLICATION
Because the man initially faces south, his first 20-metre walk produces a 20-metre southward displacement.
A left turn while facing south points east, so the next displacement is 50 metres east.
A right turn while facing east points south, so the final 80-metre walk adds to the earlier southward movement. The net displacement is 50 metres east and 100 metres south.
Therefore, d2 = (50 m)2 + (100 m)2 = 2,500 m2 + 10,000 m2 = 12,500 m2, so d = √12,500 m = 50√5 m.
CROSS-CHECK
Squaring 50√5 m gives 2,500 × 5 = 12,500 m2, which equals the sum of the squares of the perpendicular displacements.
RESULT
The distance from the front door is 50√5 m.