If `V` be the volume of a tetrahedron and `V '` be the volume of another tetrahedran formed by the centroids of faces of the previous tetrahedron and `V=K V^(prime),t h e nK` is equal to a. `9` b. `12` c. `27` d. `81`

Options

A. 9
B. 12
C. 27
D. 81

Explanation

To find the value of \( K \) in the relationship \( V = K V' \), where \( V \) is the volume of a tetrahedron and \( V' \) is the volume of another tetrahedron formed by the centroids of the faces of the original tetrahedron, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Tetrahedron**:
Consider a tetrahedron with vertices at the following coordinates:
- \( O(0, 0, 0) \)
- \( A(a, 0, 0) \)
- \( B(0, b, 0) \)
- \( C(0, 0, c) \)

2. **Volume of the Original Tetrahedron**:
The volume \( V \) of the tetrahedron can be calculated using the formula:
\[ V = \frac{1}{6} \left| \vec{OA} \cdot (\vec{OB} \times \vec{OC}) \right| \]
Here, \( \vec{OA} = (a, 0, 0) \), \( \vec{OB} = (0, b, 0) \), and \( \vec{OC} = (0, 0, c) \).
Thus, we have:
\[ V = \frac{1}{6} \cdot a \cdot b \cdot c \]

3. **Find the Centroids of the Faces**:
The centroids of the faces of the tetrahedron are calculated as follows:
- Centroid \( G_1 \) of face \( OAB \):
\[ G_1 = \left( \frac{0 + a + 0}{3}, \frac{0 + 0 + b}{3}, 0 \right) = \left( \frac{a}{3}, \frac{b}{3}, 0 \right) \]
- Centroid \( G_2 \) of face \( OAC \):
\[ G_2 = \left( \frac{0 + a + 0}{3}, 0, \frac{0 + 0 + c}{3} \right) = \left( \frac{a}{3}, 0, \frac{c}{3} \right) \]
- Centroid \( G_3 \) of face \( OBC \):
\[ G_3 = \left( 0, \frac{b}{3}, \frac{0 + 0 + c}{3} \right) = \left( 0, \frac{b}{3}, \frac{c}{3} \right) \]
- Centroid \( G_4 \) of face \( ABC \):
\[ G_4 = \left( \frac{a + 0 + 0}{3}, \frac{0 + b + 0}{3}, \frac{0 + 0 + c}{3} \right) = \left( \frac{a}{3}, \frac{b}{3}, \frac{c}{3} \right) \]

4. **Volume of the New Tetrahedron**:
The volume \( V' \) of the tetrahedron formed by the centroids \( G_1, G_2, G_3, G_4 \) can be calculated as:
\[ V' = \frac{1}{6} \left| \vec{G_1G_2} \cdot (\vec{G_1G_3} \times \vec{G_1G_4}) \right| \]
Each vector can be expressed in terms of \( a, b, c \) divided by 3:
\[ V' = \frac{1}{6} \cdot \frac{a}{3} \cdot \frac{b}{3} \cdot \frac{c}{3} = \frac{1}{6} \cdot \frac{abc}{27} = \frac{abc}{162} \]

5. **Relate the Volumes**:
From the volumes calculated, we have:
\[ V = \frac{abc}{6}, \quad V' = \frac{abc}{162} \]
Now, substituting into the equation \( V = K V' \):
\[ \frac{abc}{6} = K \cdot \frac{abc}{162} \]
Dividing both sides by \( abc \) (assuming \( abc \neq 0 \)):
\[ \frac{1}{6} = K \cdot \frac{1}{162} \]
Therefore, solving for \( K \):
\[ K = \frac{162}{6} = 27 \]

### Final Answer:
Thus, the value of \( K \) is \( 27 \).