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sindhuja P Grade: 11
        

Let ABCD be a tetrahedron with AB=41,BC=36,CA=7,DA=18,DB=27,DC=13.Find the distance between the mid points of edges AB and CD ?

7 years ago

Answers : (1)

Badiuddin askIITians.ismu Expert
147 Points
										

Dear sindhuja


In any triangle ABC having sides a,b,c Median AD is given as


AD2 = 1/4  ( 2b2 +2c2-a2)


so


let mid point of AB is E and mid point CD is F


so


in triangle ABC


 CE2 = 1/4( 2*362 + 2*72 -412)


in triangle ADB


 DE2 = 1/4( 2*272 + 2*182 -412)


and  in triangle CDE


EF2 = 1/4( 2*CE2 +2*DE2 - 132)


put the value


EF =√137


 


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Badiuddin

7 years ago
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