Quick reminder (Pythagorean theorem)
In a right triangle with legs a and b and hypotenuse c, c = √(a² + b²). We use this whenever we have a straight-line distance from two perpendicular displacements (east/west and north/south).
Problem 1
W swims: 60 miles north, 30 miles east, 30 miles north, 150 miles west.
- Combine north/south moves: 60 north + 30 north = 90 miles north.
- Combine east/west moves: 30 east and 150 west → net 30 - 150 = -120 miles (that means 120 miles west).
- So the net displacement from the start is 120 miles west and 90 miles north. These are perpendicular, so use Pythagorean theorem:
distance = √(120² + 90²) = √(14400 + 8100) = √(22500) = 150 miles.
Answer (1): 150 miles from the starting point.
Problem 2
We set up coordinates: let east be the positive x-direction and north be the positive y-direction. Place B at (0,0).
- A is 50 m east of B → A = (50, 0).
- A is 30 m west of C → C is 30 m east of A → C = (50 + 30, 0) = (80, 0).
- D is 60 m east of C → D = (80 + 60, 0) = (140, 0).
- D is 40 m east of E → D = E + (40,0), so E = D - (40,0) → E = (140 - 40, 0) = (100, 0).
- F is 50 m north of E → F = (100, 50).
- F is 80 m north of G → so G is 80 m south of F → G = (100, 50 - 80) = (100, -30).
We need distance between B(0,0) and G(100,-30). The horizontal difference is 100 m and the vertical difference is -30 m (30 m south). Use Pythagorean theorem:
distance = √(100² + (-30)²) = √(10000 + 900) = √(10900) ≈ 104.403... m.
Answer (2): To the nearest tenth, B and G are 104.4 m apart.
If you want, I can draw a simple coordinate diagram or give a quick practice problem similar to these.