Let $a = B C, b = C A, c = A B$ be the sides of the triangle $A B C$ . If $G$ is the centroid of $△ A B C$ such that $¯ ¯¯¯¯¯¯ ¯ G B$ and $¯ ¯¯¯¯¯¯ ¯ G C$ are inclined at an obtuse angle, then

5 a^{2} > b^{2} + c^{2}

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5 c^{2} > a^{2} + b^{2}

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5 b^{2} > a^{2} + c^{2}

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5 a^{2} > b^{2} - c^{2}

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Solution

The correct option is A $5 a^{2} > b^{2} + c^{2}$
$G = \frac{\to b + \to c}{3}$
$- - \to G B = \frac{\to b + \to c}{3} - \to b$ $= \frac{\to c - 2 \to b}{3}$
$- - \to G C = \frac{\to b - 2 \to c}{3}$

and,
$- - \to G B . - - \to G C < 0$
$⎛ ⎝ \frac{\to c - 2 \to b}{3} ⎞ ⎠ ⎛ ⎝ \frac{\to b - 2 \to c}{3} ⎞ ⎠ < 0$
$\to c . \to b - 2 {(c)}^{2} - 2 {(b)}^{2} + 4 \to b . \to c < 0$
$5 \to b . \to c < 2 (b^{2} + c^{2})$

and,
$∣ ∣ ∣ - - \to B C ∣ ∣ ∣ = a$
$∣ ∣ ∣ \to c - \to b ∣ ∣ ∣ = a$
$c^{2} + b^{2} - 2 \to c . \to b = a^{2}$
$c^{2} + b^{2} - a^{2} = 2 \to c . \to b$
$\frac{1}{2} (c^{2} + b^{2} - a^{2}) = \to b . \to c$

then,
$5 \to b . \to c < 2 (b^{2} + c^{2})$
$5. \frac{1}{2} (c^{2} + b^{2} - a^{2}) < 2 (b^{2} + c^{2})$
$5 a^{2} > b^{2} + c^{2}$
Hence $,$ Option $(A)$ is correct $.$

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