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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 ?

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 ?

Grade:11

1 Answers

Badiuddin askIITians.ismu Expert
147 Points
11 years ago

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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