**Displacement of a Person**

To find the displacement of a person from the origin, we need to determine the net displacement by considering the individual movements in different directions. Let's break down the given movements step by step and calculate the final displacement.

1. **Moving 30m North:**
The person moves 30m in the north direction. Since this movement is in a straight line, the displacement is also 30m north.

2. **Moving 20m East:**
After moving north, the person then moves 20m towards the east direction. Again, this movement is in a straight line, so the displacement is 20m east.

3. **Moving 30√2m in South West direction:**
Finally, the person moves 30√2m in the South West direction. South West is the combination of South and West. Since the person moves 30√2m at an angle of 45 degrees, we can break it down into its components using trigonometry.

The horizontal component (West) is given by 30√2 * cos(45°) ≈ 30m * 0.707 ≈ 21.21m West.
The vertical component (South) is given by 30√2 * sin(45°) ≈ 30m * 0.707 ≈ 21.21m South.

Therefore, the displacement due to this movement is 21.21m South and 21.21m West.

**Calculating the Net Displacement:**

To calculate the net displacement, we need to consider the vector sum of all the individual displacements.

The displacement in the north direction is 30m.
The displacement in the east direction is 20m.
The displacement in the south direction is 21.21m.
The displacement in the west direction is 21.21m.

Using the Pythagorean theorem, we can find the magnitude of the net displacement:

Net Displacement^2 = (30m - 21.21m)^2 + (20m - 21.21m)^2
Net Displacement^2 = 8.79m^2 + (-1.21m)^2
Net Displacement^2 = 8.79m^2 + 1.46m^2
Net Displacement^2 = 10.25m^2

Taking the square root of both sides, we get:

Net Displacement ≈ √10.25m^2
Net Displacement ≈ 3.2m

Therefore, the net displacement of the person from the origin is approximately 3.2m. The direction can be described as 8.79m South and 1.21m West in a vector form.

10 m along west