The distance between the origin and the centroid of the tetrahedron whose vertices are $(0, 0, 0)$ , $(a, 0, 0), (0, b, 0), (0, 0, c)$ is?

\sqrt{a^{2} + b^{2} + c^{2}}

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\frac{\sqrt{a^{2} + b^{2} + c^{2}}}{2}

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\frac{\sqrt{a^{2} + b^{2} + c^{2}}}{4}

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4 \sqrt{a^{2} + b^{2} + c^{2}}

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Solution

The correct option is C $\frac{\sqrt{a^{2} + b^{2} + c^{2}}}{4}$
Given,
Centroid of tatrahedron = $(\frac{a}{4}, \frac{b}{4}, \frac{c}{4})$
distance between origin & centroid of tetrahedron..................
$(\frac{a}{4}, \frac{b}{4}, \frac{c}{4})$ (0,0,0)
$\begin{matrix} = \sqrt{{(\frac{a}{4})}^{2} + {(\frac{b}{4})}^{2} + {(\frac{c}{4})}^{2}} = \frac{\sqrt{a_{2} + b_{2} + c_{2}}}{4} \end{matrix}$
So that the correct option is C.

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Similar questions

If a + b + c = 2s, then prove the following identities

(a) s² + (s − a)² + (s − b)² + (s − c)² = a² + b² + c²

(b) a² + b² − c² + 2ab = 4s (s − c)

(d) a² − b² − c² + 2ab = 4(s − b) (s − c)

(e) (2bc + a² − b² − c²) (2bc − a² + b² + c²) = 16s (s − a) (s − b) (s − c)

(f)

|\begin{matrix} - a (b^{2} + c^{2} - a^{2}) & 2 b^{3} & 2 c^{3} \\ 2 a^{3} & - b (c^{2} + a^{2} - b^{2}) & 2 c^{3} \\ 2 a^{3} & 2 b^{3} & - c (a^{2} + b^{2} - c^{2}) \end{matrix}| = a b c {(a^{2} + b^{2} + c^{2})}^{3}

\frac{a^{2} - {(b - c)}^{2}}{{(a + c)}^{2} - b^{2}} + \frac{b^{2} - {(a - c)}^{2}}{{(a + b)}^{2} - c^{2}} + \frac{c^{2} - {(a - b)}^{2}}{{(b + c)}^{2} - a^{2}}

\frac{a^{2} - b^{2} - c^{2}}{(a - b) (a - c)} + \frac{b^{2} - c^{2} - a^{2}}{(b - c) (b - a)} + \frac{c^{2} - a^{2} - b^{2}}{(c - a) (c - b)}

If a, b, c, d are in G.p., prove that :

(i) $(a^{2} + b^{2}), (b^{2} + c^{2}), (c^{2} + d)^{2}$ are in G.P.

(ii) $(a^{2} - b^{2}), (b^{2} - c^{2}), (c^{2} - d)^{2}$ are in G.P.

(iii) $\frac{1}{a^{2} + b^{2}}, \frac{1}{b^{2} + c^{2}}, \frac{1}{c^{2} + d^{2}}$ are in G.P.

(iv) $(a^{2} + b^{2} + c^{2}), (a b + b c + c d), (b^{2} + c^{2} + d^{2})$

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