The volume of a regular tetrahedron is V=(a^3)/(6√2) , where a= 2 cm is the measure of
each of the six edges. The error of the measue of the edge is ϵ= 0.2 mm.
evaluate the relative error of measurement

To find the relative error of the volume measurement, we'll follow these steps:

1. **Express the volume \( V \) of the tetrahedron**:
   \[
   V = \frac{a^3}{6\sqrt{2}}
   \]
   where \( a = 2 \) cm is the edge length.

2. **Identify the absolute error of \( a \)**:
   Given that \( \epsilon = 0.2 \) mm = 0.02 cm, this is the absolute error of the edge length.

3. **Calculate the relative error in the edge length \( a \)**:
   \[
   \text{Relative error in } a = \frac{\epsilon}{a} = \frac{0.02 \text{ cm}}{2 \text{ cm}} = 0.01
   \]

4. **Determine how the error propagates to the volume \( V \)**:
   Since \( V \) depends on \( a^3 \), the relative error in \( V \) will be three times the relative error in \( a \):
   \[
   \text{Relative error in } V = 3 \times \text{Relative error in } a = 3 \times 0.01 = 0.03
   \]

Therefore, the **relative error of the volume measurement** is \( 0.03 \) or **3%**.

Would you like more details or have any questions?

Here are 5 related questions to expand your understanding:
1. How does the relative error change if the edge length error changes?
2. What is the absolute error in the volume of the tetrahedron?
3. Can you calculate the exact volume of the tetrahedron using the given values?
4. How would the formula change if the shape was a regular hexahedron (cube)?
5. What is the significance of the relative error in practical measurements?

**Tip:** When calculating relative errors, remember that for quantities raised to a power \( n \), the relative error is multiplied by \( n \).

The volume of a regular tetrahedron is V=(a^3)/(6√2), where a= 2 cm is the measure of each of the six edges. The error of the measure of the edge is ϵ= 0.2 mm. Evaluate the relative error of measurement.

Learn how to calculate the relative error in the volume of a regular tetrahedron with edge length a = 2 cm and measurement error of ϵ = 0.2 mm. This problem involves error propagation and geometry concepts, suitable for advanced high school students.

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