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Polyhedrons

Let V be the ...

Question

Let $V$ be the volume of a tetrahedron and $V^{'}$ be the volume of another tetrahedron formed by the centroids of the previous tetrahedron. If $V = k V^{'}$ , then the value of $k$ is

27.0

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27.00

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Solution

Let $O A B C$ be the original tetrahedron.
$V = \frac{1}{6} [\to a \to b \to c]$
Let $G, G_{1}, G_{2}, G_{3}$ be the centroids of the original tetrahedron.
$\to G = \frac{\to a + \to b + \to c}{3}$
$_{1} = \frac{\to a + \to b}{3};_{2} = \frac{\to b + \to c}{3};_{3} = \frac{\to a + \to c}{3}$
$- -- \to G_{1} G = \frac{\to c}{3}; - -- \to G_{2} G = \frac{\to a}{3}; - -- \to G_{3} G = \frac{\to b}{3}$

$V^{'} =$ Volume of tetrahedron $G G_{1} G_{2} G_{3}$
$= \frac{1}{6} [- -- \to G G_{1} - -- \to G G_{2} - -- \to G G_{3}] = \frac{1}{6} ⎡ ⎣ \frac{\to a}{3} \frac{\to b}{3} \frac{\to c}{3} ⎤ ⎦ = \frac{1}{6} \cdot \frac{1}{27} [\to a \to b \to c] V^{'} = \frac{V}{27}$
$\Rightarrow V = 27 V^{'}$
$\Rightarrow k = 27$

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