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To solve the problem, we need to calculate the area of the circle inscribed in a right triangle with legs (catetos) of 6 meters and 8 meters.

### Step-by-Step Solution:

1. **Find the Hypotenuse (c):**
   The triangle is a right triangle, so we can apply the Pythagorean theorem:

   \[
   c = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, \text{m}
   \]

2. **Find the Semi-Perimeter (s):**
   The semi-perimeter \(s\) is half of the perimeter of the triangle, which is the sum of all sides:
   
   \[
   s = \frac{6 + 8 + 10}{2} = \frac{24}{2} = 12 \, \text{m}
   \]

3. **Inradius (r):**
   The formula for the inradius \(r\) of a right triangle is:

   \[
   r = \frac{a + b - c}{2}
   \]
   where \(a = 6 \, \text{m}, b = 8 \, \text{m},\) and \(c = 10 \, \text{m}\).

   So:

   \[
   r = \frac{6 + 8 - 10}{2} = \frac{4}{2} = 2 \, \text{m}
   \]

4. **Area of the Inscribed Circle:**
   The area \(A\) of a circle is given by the formula:

   \[
   A = \pi r^2
   \]
   Substituting the value of \(r = 2 \, \text{m}\):

   \[
   A = \pi (2^2) = 4\pi \, \text{m}^2
   \]

Thus, the area of the inscribed circle is \( \boxed{4\pi \, \text{m}^2} \).

The correct answer is **B: \(4\pi \, \text{m}^2\)**.

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Do you want more details or have any questions?

Here are 5 related questions:
1. How is the formula for the inradius of a right triangle derived?
2. What is the Pythagorean theorem, and how is it applied to right triangles?
3. How do you calculate the perimeter and semi-perimeter of any triangle?
4. What is the geometric significance of the inradius in a triangle?
5. How can the inscribed circle be applied to non-right triangles?

**Tip:** The Pythagorean theorem only applies to right triangles; for other triangles, you must use the Law of Cosines.

Calcula el área de un círculo inscrito en un triángulo rectángulo cuyos catetos miden 6 m y 8 m.

Learn how to calculate the area of a circle inscribed in a right triangle with legs of 6 m and 8 m. This problem applies key geometric concepts such as the Pythagorean theorem and the formula for the inradius. Suitable for students in Grades 9-12.

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