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Thread: 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 $\displaystyle CD$ is perpendicular to $\displaystyle AB$.

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

    $\displaystyle ADB=3:2 $

    $\displaystyle \Rightarrow AD:AB = 3: 5 $

    $\displaystyle \Rightarrow AD = \tfrac35AB = 39$

    $\displaystyle \Rightarrow BD = 26$

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

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

    $\displaystyle \Rightarrow AC = 13\sqrt{13}$

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

    $\displaystyle \Rightarrow BC = 26\sqrt2$

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

    Grandad
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  3. #3
    RAz
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    Thank you so much!
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