Question

in the given figure ABCD is a rectangle if AB =20cm,BC =12 cm and AF:FC =3:2 then find the length of FE in centemeters

Answer 1

I believe there is some typing mistake some where... 
I think that:  you should mention that:         AF : FE = 3 : 2 
then it is simpler..   

         Let  AF = 3 x  and  FE = 2 x

Draw a perpendicular from F on to AB to meet that at G.  Use similar triangles principles.

 EC / EF  = FG / FA
   EC / 2x  = 12 / 3x           =>  EC  = 8 cm

So :   BE = 12 + 8 = 20 cm        and  AB = 20 cm.
  So  AE = 20√2  cm

FE =  AE * 2/5  = 8√2 cm
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In case  it is not a mistake and it is only AF : FC =  3 : 2  then,

FG = BC = 12 cm
AG = AB - BG =  AB - FC =  20 - 2 x  cm

Pythagoras theorem in triangle AFG,  we have:
       9 x²  = 12² + (20 - 2x)²
  =>  x² + 16 x - 108.8 = 0
 =>  x = √172.8  - 8  cm    ≈ 5.145  cm

AF = 3x   ≈  15.435  cm
AG = 20 - 2x =  9.71  cm
FC = 2x = 10.29 cm

 FE   ≈  16.36 cm