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Math Help - Sides of a Quadrilateral

  1. #1
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    Sides of a Quadrilateral

    Hello everyone. Iím working on the following problem involving a geometric progression and a quadrilateral:
    Problem:
    The lengths of the sides of a quadrilateral are in geometric progression and the longest side is 81 cm. Given that the perimeter is 120 cm, find the lengths of the other three sides.
    Hereís what Iíve done with this problem so far:
    Let the lengths of the four sides of the quadrilateral be denoted by t_1 < t_2 < t_3 < t_4.
    t_4 = 81 cm (Given)
    t_1 + t_2 + t_3 + t_4 = 120 cm (Given Perimeter of Quadrilateral)
    Substituting for t_4 above:
    t_1 + t_2 + t_3 + 81 = 120
    t_1 + t_2 + t_3 = 39
    a + ar + ar^2 = 39
    a ( 1 + r + r^2 ) = 39
    This tells us that the sum of the lengths of the other three sides equals 39 but we need to find the lengths for all three. How to proceed from here on...?? Need to basically find a and r.
    Any and all help will be greatly appreciated. Iím a little rusty at this after many years of keeping away from math.
    Thank you.
    Shahz.
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  2. #2
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    geometric progression/ quad

    Quote Originally Posted by pollardrho06 View Post
    Hello everyone. Iím working on the following problem involving a geometric progression and a quadrilateral:
    Problem:
    The lengths of the sides of a quadrilateral are in geometric progression and the longest side is 81 cm. Given that the perimeter is 120 cm, find the lengths of the other three sides.
    Hereís what Iíve done with this problem so far:
    Let the lengths of the four sides of the quadrilateral be denoted by t_1 < t_2 < t_3 < t_4.
    t_4 = 81 cm (Given)
    t_1 + t_2 + t_3 + t_4 = 120 cm (Given Perimeter of Quadrilateral)
    Substituting for t_4 above:
    t_1 + t_2 + t_3 + 81 = 120
    t_1 + t_2 + t_3 = 39
    a + ar + ar^2 = 39
    a ( 1 + r + r^2 ) = 39
    This tells us that the sum of the lengths of the other three sides equals 39 but we need to find the lengths for all three. How to proceed from here on...?? Need to basically find a and r.
    Any and all help will be greatly appreciated. Iím a little rusty at this after many years of keeping away from math.
    Thank you.
    Shahz.
    Hello Shahz,
    The progression must start with an integer. 1 works.Easy going after that


    bjh
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  3. #3
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    Quote Originally Posted by bjhopper View Post
    Hello Shahz,
    The progression must start with an integer. 1 works.Easy going after that


    bjh
    Thanks for responding bjh.

    Why are you taking the first term to be 1? What if it isn't 1? I'm not sure I see what you're trying to do here. What am I missing?
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  4. #4
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    Quote Originally Posted by pollardrho06 View Post
    Hello everyone. Iím working on the following problem involving a geometric progression and a quadrilateral:
    Problem:
    The lengths of the sides of a quadrilateral are in geometric progression and the longest side is 81 cm. Given that the perimeter is 120 cm, find the lengths of the other three sides.
    Hereís what Iíve done with this problem so far:
    Let the lengths of the four sides of the quadrilateral be denoted by t_1 < t_2 < t_3 < t_4.
    t_4 = 81 cm (Given)
    t_1 + t_2 + t_3 + t_4 = 120 cm (Given Perimeter of Quadrilateral)
    Substituting for t_4 above:
    t_1 + t_2 + t_3 + 81 = 120
    t_1 + t_2 + t_3 = 39
    a + ar + ar^2 = 39
    a ( 1 + r + r^2 ) = 39
    This tells us that the sum of the lengths of the other three sides equals 39 but we need to find the lengths for all three. How to proceed from here on...?? Need to basically find a and r.

    \color{blue}a+ar+ar^2+ar^3=120

    \color{blue}ar^3=81

    \color{blue}a+ar+ar^2=39

    \color{blue}\frac{ar^3}{a\left(1+r+r^2\right)}=\fr  ac{81}{39}=\frac{3(27)}{3(13)}

    It should be clearer now

    Any and all help will be greatly appreciated. Iím a little rusty at this after many years of keeping away from math.
    Thank you.
    Shahz.
    .
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  5. #5
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    geometric progression/quad

    [quote=pollardrho06;516247]Thanks for responding bjh.

    Why are you taking the first term to be 1? What if it isn't 1? I'm not sure I see what you're trying to do here. What am I missing?[/quote


    Hi shahz,

    Iknew that 81 is 3^4 so Imade r=3 and a =1

    Making the progression 1,3,9,27,81 but since 81 is the fourth side the other 3 are 3,9,27. Dig?

    bjh
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  6. #6
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    [quote=bjhopper;516333]
    Quote Originally Posted by pollardrho06 View Post
    Thanks for responding bjh.

    Why are you taking the first term to be 1? What if it isn't 1? I'm not sure I see what you're trying to do here. What am I missing?[/quote


    Hi shahz,

    Iknew that 81 is 3^4 so Imade r=3 and a =1

    Making the progression 1,3,9,27,81 but since 81 is the fourth side the other 3 are 3,9,27. Dig?

    bjh
    Nice!! Thanks a bunch mate.
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  7. #7
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    Quote Originally Posted by pollardrho06 View Post
    Hello everyone. Iím working on the following problem involving a geometric progression and a quadrilateral:
    Problem:
    The lengths of the sides of a quadrilateral are in geometric progression and the longest side is 81 cm. Given that the perimeter is 120 cm, find the lengths of the other three sides.
    Hereís what Iíve done with this problem so far:
    Let the lengths of the four sides of the quadrilateral be denoted by t_1 < t_2 < t_3 < t_4.
    t_4 = 81 cm (Given)
    t_1 + t_2 + t_3 + t_4 = 120 cm (Given Perimeter of Quadrilateral)
    Substituting for t_4 above:
    t_1 + t_2 + t_3 + 81 = 120
    t_1 + t_2 + t_3 = 39
    a + ar + ar^2 = 39
    a ( 1 + r + r^2 ) = 39
    This tells us that the sum of the lengths of the other three sides equals 39 but we need to find the lengths for all three. How to proceed from here on...?? Need to basically find a and r.
    Any and all help will be greatly appreciated. Iím a little rusty at this after many years of keeping away from math.
    Thank you.
    Shahz.
    Alternatively,

    for a geometric series,

    S_n=\frac{a\left(1-r^n\right)}{1-r}=\frac{a\left(r^n-1\right)}{r-1}

    The sum of the first 4 terms is 120

    S_4=\frac{a\left(r^4-1\right)}{r-1}

    ar^3=81\ \Rightarrow\ S_3=a+ar+ar^2=S_4-ar^3=120-81=39

    S_3=\frac{a\left(r^3-1\right)}{r-1}

    \frac{S_4}{S_3}=\frac{120}{39}=S_4\left(\frac{1}{S  _3}\right)=\frac{r^4-1}{r^3-1}

    \left(r^4-1\right)39=\left(r^3-1\right)120

    39r^4-39=120r^3-120

    120r^3-39r^4=120-39=81

    r^3(120-39r)=81

    3r^3(40-13r)=3(27)

    r^3(40-13r)=27=3^3

    Hence r=3 is a solution as this gives 3^3(40-39)=3^3
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  8. #8
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    Quote Originally Posted by Archie Meade View Post
    Alternatively,

    for a geometric series,

    S_n=\frac{a\left(1-r^n\right)}{1-r}=\frac{a\left(r^n-1\right)}{r-1}

    The sum of the first 4 terms is 120

    S_4=\frac{a\left(r^4-1\right)}{r-1}

    ar^3=81\ \Rightarrow\ S_3=a+ar+ar^2=S_4-ar^3=120-81=39

    S_3=\frac{a\left(r^3-1\right)}{r-1}

    \frac{S_4}{S_3}=\frac{120}{39}=S_4\left(\frac{1}{S  _3}\right)=\frac{r^4-1}{r^3-1}

    \left(r^4-1\right)39=\left(r^3-1\right)120

    39r^4-39=120r^3-120

    120r^3-39r^4=120-39=81

    r^3(120-39r)=81

    3r^3(40-13r)=3(27)

    r^3(40-13r)=27=3^3

    Hence r=3 is a solution as this gives 3^3(40-39)=3^3

    Lovely!! You guys just made my day!! God bless you...
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