Geometry-8th gr

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The perimeter of triangle ABC is 120. LIne segment BD bisects <B. Line segment AD= 15, and line segment DC= 21. FInd AB and BC.


Thanks.

  • Geometry-8th gr -

    This is a very tough question for grade 8, I am totally impressed.

    make your sketch, let
    AB = a
    BC = b
    BD = x
    Angle B = 2Ø , so that angle ABD = Ø and angle BCD = Ø

    I am going to use the fact that if we have triangle with sides a and b with Ø the angle between them
    the area = (1/2)ab sinØ

    area of triangle ABD= (1/2)ax sinØ
    area of triangle BCD=(1/2)bx sinØ

    But if we consider the bases along AC , the both have the same measured from B
    so they are in the ratio of 15:21 or 5:7

    then:
    (1/2)axsinØ/( (1/2)bxsinØ ) = 5/7
    canceling reduces this to
    a/b = 5/7
    5b = 7a
    b = 7a/5 = 1.4a ---- ******

    now let's use the perimeter:
    a + b + 15+21 = 120
    a+b =84 -----******

    sub in b = 1.4a into the above equation
    a+b = 84
    a + 1.4a = 84
    2.4a = 84
    a = 84/2.4 = 35
    then b = 1.4(35) = 49


    great question !!!!

  • Geometry-8th gr -

    or, using the angle bisector theorem,

    AD/DC = AB/BC
    so, as Reiny calculated,

    AB/BC = 5/7
    AD+BC = 84
    5x+12x=84
    x=7

    so, AB = 35
    BC = 49

  • Geometry-8th gr - typo -

    oops
    5x+7x=84
    x=7

  • Geometry-8th gr -

    Thanks.

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