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$If the sides of △ A B C are 5, 7, 8 units then A G^{2} + B G^{2} + C G^{2} =$

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Detailed Solution

${AB}^{2} + {BC}^{2} + {CA}^{2} = 3 ({GA}^{2} + {GB}^{2} + {GC}^{2})$

$\Rightarrow 25 + 49 + 64 = 3 (G A^{2} + G B^{2} + G C^{2})$

$\Rightarrow \frac{138}{3} = G A^{2} + G B^{2} + G C^{2}$

$\Rightarrow 46 = G A^{2} + G B^{2} + G C^{2}$

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