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Centroid of a Triangle

Let ABCD be t...

Question

Let ABCD be tetrahedron with AB = 41, AC = 7, AD = 18, BC = 36, BD = 27 and CD = 13 as shown in figure. Let d be the distance between the midpoints of edges AB and CD, then $[\frac{d}{2}]$ is
(where [.] denotes greatest integer functions) ___

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Solution

Let ‘x’ be the median from A to CD and y be the median from B to CD.

$∴ 2 (7^{2} + 18^{2}) = 4 x^{2} + 13^{2} & 2 (27^{2} + 36^{2}) = 4 y^{2} + 13^{2} & 2 (x^{2} + y^{2}) = 41^{2} + 4 d^{2} = 7^{2} + 8^{2} + 27^{2} + 36^{2} - 13^{2} ∴ d^{2} = 137$

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