If the lengths of sides of $△ A B C$ are $5, 7, 8$ units then $A G^{2} + B G^{2} + C G^{2} =$
(where $G$ is the centroid)

46

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138

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92

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69

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Solution

The correct option is A $46$

We know that
$3 (A B^{2} + B C^{2} + A C^{2}) = 4 (A D^{2} + B E^{2} + C F^{2})$
Also, $A G = \frac{2}{3} A D$
$\Rightarrow A D^{2} + B E^{2} + C F^{2} = \frac{9}{4} (A G^{2} + B G^{2} + C G^{2})$
$∴ A B^{2} + B C^{2} + A C^{2} = \frac{4}{3} \times \frac{9}{4} (A G^{2} + B G^{2} + C G^{2})$
$\Rightarrow O G^{2} + B G^{2} + C G^{2} = \frac{1}{3} (A B^{2} + B C^{2} + A C^{2})$
$= \frac{1}{3} (5^{2} + 7^{2} + 8^{2}) = \frac{138}{3} = 46$

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