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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

Please feel free to post as many doubts on our discussion forum as you can.If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very  quickly. We are all IITians and here to help you in your IIT JEE  & AIEEE preparation. All the best. Regards,Askiitians ExpertsBadiuddin
```
7 years ago
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