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 [d2] is
(where [.] denotes greatest integer functions)
Let ‘x’ be the median from A to CD and y be the median from B to CD.
∴ 2(72+182)=4x2+132& 2(272+362)=4y2+132& 2(x2+y2)=412+4d2=72+82+272+362−132∴ d2=137