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 ?

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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