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Math Help - Pinding perimeter of triangle in surd form

  1. #1
    RAz
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    Pinding perimeter of triangle in surd form

    The points A and B lie on the line y=12x / 5
    The point D divides the interval AB into the ratio AD : DB = 3 : 2
    C is the point (24, -10)
    Area of triangle is 845 square units.

    I have found that

    CD = 26 (I am assuming D is perpendicular C)

    by using | ax + by + c| / sqrt(a^2 + b^2) [I first put the line equitation into general form then plugged away]

    AB = 65

    From 1/2 x AB x CD = 845

    I do not know how to find the perimeter of this triangle, as shown below:


    Any help would be great. Exam tomorrow
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  2. #2
    MHF Contributor
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    Hello RAz
    Quote Originally Posted by RAz View Post
    The points A and B lie on the line y=12x / 5
    The point D divides the interval AB into the ratio AD : DB = 3 : 2
    C is the point (24, -10)
    Area of triangle is 845 square units.

    I have found that

    CD = 26 (I am assuming D is perpendicular C)

    by using | ax + by + c| / sqrt(a^2 + b^2) [I first put the line equitation into general form then plugged away]

    AB = 65

    From 1/2 x AB x CD = 845

    I do not know how to find the perimeter of this triangle, as shown below:


    Any help would be great. Exam tomorrow
    Your working is fine so far, on the assumption that CD is perpendicular to AB.

    You now need to find the distances AD and BD, which you can do like this:

    B=3:2 " alt="ADB=3:2 " />

    \Rightarrow AD:AB = 3: 5

    \Rightarrow AD = \tfrac35AB = 39

    \Rightarrow BD = 26

    Now use Pythagoras' Theorem on \triangle ADC: AC^2 = 39^2+26^2

    = 13^2(3^2+2^2)=13^2\times13

    \Rightarrow AC = 13\sqrt{13}

    and on \triangle BDC: BC^2=26^2+26^2=26^2\times 2

    \Rightarrow BC = 26\sqrt2

    So the perimeter of \triangle ABC = 65+13\sqrt{13}+26\sqrt2 = 13(5+\sqrt{13}+2\sqrt2)

    Grandad
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  3. #3
    RAz
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    Thank you so much!
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