Right triangle ABC with sides 3m, 4m, and 5m, demonstrating the Pythagorean theorem.

The image shows a geometric path involving three points: A, B, and C, with distances marked between them.

Given:

- From point A to B: 4 meters (upward arrow)
- From point B to C: 3 meters (rightward arrow)
- From point A to C: 5 meters (diagonal arrow)

We are to analyze this triangle formed by points A, B, and C.

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Step 1: Identify the Triangle

We have a triangle ABC with sides:
- AB = 4 m
- BC = 3 m
- AC = 5 m

Let’s check if this is a right triangle using the Pythagorean Theorem:

> $$
> \text{If } AB^2 + BC^2 = AC^2, \text{ then it's a right triangle at } B.
> $$

Calculate:
- $ AB^2 = 4^2 = 16 $
- $ BC^2 = 3^2 = 9 $
- Sum: $ 16 + 9 = 25 $
- $ AC^2 = 5^2 = 25 $

✔ Since $ AB^2 + BC^2 = AC^2 $, the triangle satisfies the Pythagorean theorem.

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Conclusion:

- Triangle ABC is a right triangle, with the right angle at point B.
- This means the path from A to B to C forms a right angle at B.
- The direct distance from A to C is the hypotenuse of the right triangle.

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Final Answer:

The triangle ABC is a right triangle with legs 3 m and 4 m, and hypotenuse 5 m. It follows the 3-4-5 Pythagorean triple.

So, the solution confirms that:
> The path from A → B → C forms a right angle at B, and the straight-line distance from A to C is 5 meters, which matches the given diagonal.

This is a classic example of the Pythagorean theorem in real-life geometry.

✔ Answer: The triangle ABC is a right triangle with a right angle at B, satisfying $ 3^2 + 4^2 = 5^2 $.