Peter leaves his house and walks 15 meters south, then turns east and walks another 15 meters. How far is he from his house? Round your answer to the nearest tenth

Use the Pythagorean Theorem.

a^2 + b^2 = c^2

15^2 + 15^2 = c^2

450 = c^2
square root 450 = 21.2

To find the distance from Peter's initial position, we can use the Pythagorean theorem.

Let's break it down step by step:

1. Peter walks 15 meters south. This creates a vertical displacement.

2. Peter then turns east and walks another 15 meters. This creates a horizontal displacement.

Since these displacements are perpendicular to each other, we can treat them as the two legs of a right triangle, with Peter's starting position as the right angle.

Now, let's find the distance from Peter's house using the Pythagorean theorem:

1. The vertical displacement (south) is 15 meters. This will be the length of one of the legs of the right triangle.

2. The horizontal displacement (east) is also 15 meters, and it will be the length of the other leg of the right triangle.

3. To find the hypotenuse (the distance from Peter's house), we use the formula:

c^2 = a^2 + b^2

where c is the hypotenuse and a and b are the lengths of the legs.

Plugging in the values:

c^2 = 15^2 + 15^2

c^2 = 225 + 225

c^2 = 450

4. Now, we can solve for c by taking the square root of both sides:

c = √450

5. Approximating the square root of 450 to the nearest tenth:

c ≈ 21.2 meters

Therefore, Peter is approximately 21.2 meters away from his house when he turns east and walks another 15 meters.