Question

If V be the volume of a tetrahedron and V′ be the volume of another tetrahedron formed by the centroids of faces of the previous tetrahedron and $V=KV'$, then K is equal to
(a)9
(b)12
(c)27
(d)81

Answer 1

Hint: In this question, first work out on the volume of the tetrahedron by using the formula$V=\dfrac{1}{6}\left[ \overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c} \right]$. Second work out on the coordinates of the centroids of the face of the tetrahedron by using formula $\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3},\dfrac{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}}{3} \right)$ .

Complete step-by-step answer:
Let $O(0,0,0),A(a,0,0),B(0,b,0)$ and $C(0,0,c)$ be the vertices of the given tetrahedron.
Let $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ be the position vectors of the points A, B and C respectively.

The volume of the given tetrahedron is given by

 $V=\dfrac{1}{6}\left[ \overrightarrow{OA}\text{ }\overrightarrow{OB}\text{ }\overrightarrow{OC} \right]=\dfrac{1}{6}\left[ \overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c} \right]................(1)$

There are four triangles in the given tetrahedron. The triangles are \[\Delta ABC,\Delta AOB,\Delta BOC\] and$\Delta AOC$.

Now the centroids of the triangles are $O(0,0,0),A\left( \dfrac{a}{3},0,0 \right),B\left( 0,\dfrac{b}{3},0 \right)$ and $C\left( 0,0,\dfrac{c}{3} \right)$ respectively.

The volume another tetrahedron formed by centroids of faces of the given tetrahedron is given by

\[V'=\dfrac{1}{6}\left[ \overrightarrow{OA}\text{ }\overrightarrow{OB}\text{ }\overrightarrow{OC} \right]=\dfrac{1}{6}\left[ \dfrac{\overrightarrow{a}}{3}\text{ }\overrightarrow{\dfrac{b}{3}}\text{ }\overrightarrow{\dfrac{c}{3}} \right]\]

\[V'=\dfrac{1}{6}\dfrac{1}{27}\left[ \overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c} \right]\]

\[V'=\dfrac{1}{27}\left( \dfrac{1}{6}\left[ \overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c} \right] \right)\]

\[V'=\dfrac{1}{27}V\] (From equation (1))

Multiplying both sides by 27, we get

\[V=27V'\]

This is comparing with$V=KV'$, we get

K = 27.

Therefore, the correct option for the given question is option ©

Note: In geometry, a tetrahedron also known as a triangular pyramid is a polyhedron composed of four triangular faces, six straight edges, and four vertex corners. The tetrahedron is the simplest of all the ordinary convex polyhedral and the only one that has fewer than 5 faces.