△ABC中,DE‖FG‖BC,S△ADE:S四边形DEGF:S四边形FGCB=4:5:16,且FD=2,则FB的长为.

3个回答

  • ∵DE//FG//BC

    ∴△ADE∽△AFG∽△ABC

    ∵S△ADE:S四边形DEGF:S四边形FGCB = 4:5:16

    ∴S△ADE:S△AFG:S△ABC = 4:(4+5):(4+5+16)

    = 4:9:16

    ∵相似三角形面积的比等于相似比的平方

    ∴(AD:AF)^2 = 4:9

    (AF:AB)^2 = 9:16

    ∴ AD:AF = 2:3

    AF:AB = 3:5

    ∴AD:(AF-AD)=2:(3-2)

    AF:(AB-AF)=3:(5-3)

    即AD=2DF=2*2=4

    BF=2AF/3=2*(4+2)/3=4

    2)延长BA,CD交于M点

    则MA:MB=AD:BC=3:9=1:3

    ∴MA:(MB-MA)=1:(3-1)

    即MA:4=1:2

    ∴MA=2

    ∴S△MAD:S△MBC=(1:3)^2=1:9