Sides of a Quadrilateral

Feb 2010
36
2
New Jersey, USA
[FONT=&quot]Hello everyone. I’m working on the following problem involving a geometric progression and a quadrilateral:[/FONT]
[FONT=&quot]Problem: [/FONT]
[FONT=&quot]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.[/FONT]
[FONT=&quot]Here’s what I’ve done with this problem so far:[/FONT]
[FONT=&quot]Let the lengths of the four sides of the quadrilateral be denoted by t_1 < t_2 < t_3 < t_4.[/FONT]
[FONT=&quot]t_4 = 81 cm (Given)[/FONT]
[FONT=&quot]t_1 + t_2 + t_3 + t_4 = 120 cm (Given Perimeter of Quadrilateral)[/FONT]
[FONT=&quot]Substituting for t_4 above:[/FONT]
[FONT=&quot]t_1 + t_2 + t_3 + 81 = 120[/FONT]
[FONT=&quot]t_1 + t_2 + t_3 = 39[/FONT]
[FONT=&quot]a + ar + ar^2 = 39[/FONT]
[FONT=&quot]a ( 1 + r + r^2 ) = 39[/FONT]
[FONT=&quot]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.[/FONT]
[FONT=&quot]Any and all help will be greatly appreciated. I’m a little rusty at this after many years of keeping away from math.[/FONT]
[FONT=&quot]Thank you.[/FONT]
[FONT=&quot]Shahz.[/FONT]
 
Nov 2007
985
175
Trumbull Ct
geometric progression/ quad

[FONT=&quot]Hello everyone. I’m working on the following problem involving a geometric progression and a quadrilateral:[/FONT]
[FONT=&quot]Problem: [/FONT]
[FONT=&quot]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.[/FONT]
[FONT=&quot]Here’s what I’ve done with this problem so far:[/FONT]
[FONT=&quot]Let the lengths of the four sides of the quadrilateral be denoted by t_1 < t_2 < t_3 < t_4.[/FONT]
[FONT=&quot]t_4 = 81 cm (Given)[/FONT]
[FONT=&quot]t_1 + t_2 + t_3 + t_4 = 120 cm (Given Perimeter of Quadrilateral)[/FONT]
[FONT=&quot]Substituting for t_4 above:[/FONT]
[FONT=&quot]t_1 + t_2 + t_3 + 81 = 120[/FONT]
[FONT=&quot]t_1 + t_2 + t_3 = 39[/FONT]
[FONT=&quot]a + ar + ar^2 = 39[/FONT]
[FONT=&quot]a ( 1 + r + r^2 ) = 39[/FONT]
[FONT=&quot]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.[/FONT]
[FONT=&quot]Any and all help will be greatly appreciated. I’m a little rusty at this after many years of keeping away from math.[/FONT]
[FONT=&quot]Thank you.[/FONT]
[FONT=&quot]Shahz.[/FONT]
Hello Shahz,
The progression must start with an integer. 1 works.Easy going after that


bjh
 
Feb 2010
36
2
New Jersey, USA
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?
 
Dec 2009
3,120
1,342
[FONT=&quot]Hello everyone. I’m working on the following problem involving a geometric progression and a quadrilateral:[/FONT]
[FONT=&quot]Problem: [/FONT]
[FONT=&quot]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.[/FONT]
[FONT=&quot]Here’s what I’ve done with this problem so far:[/FONT]
[FONT=&quot]Let the lengths of the four sides of the quadrilateral be denoted by t_1 < t_2 < t_3 < t_4.[/FONT]
[FONT=&quot]t_4 = 81 cm (Given)[/FONT]
[FONT=&quot]t_1 + t_2 + t_3 + t_4 = 120 cm (Given Perimeter of Quadrilateral)[/FONT]
[FONT=&quot]Substituting for t_4 above:[/FONT]
[FONT=&quot]t_1 + t_2 + t_3 + 81 = 120[/FONT]
[FONT=&quot]t_1 + t_2 + t_3 = 39[/FONT]
[FONT=&quot]a + ar + ar^2 = 39[/FONT]
[FONT=&quot]a ( 1 + r + r^2 ) = 39[/FONT]
[FONT=&quot]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.[/FONT]

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

\(\displaystyle \color{blue}ar^3=81\)

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

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

It should be clearer now

[FONT=&quot]Any and all help will be greatly appreciated. I’m a little rusty at this after many years of keeping away from math.[/FONT]
[FONT=&quot]Thank you.[/FONT]
[FONT=&quot]Shahz.[/FONT]
.
 
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Reactions: pollardrho06
Nov 2007
985
175
Trumbull Ct
geometric progression/quad

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
 
  • Like
Reactions: pollardrho06
Feb 2010
36
2
New Jersey, USA
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.
 
Dec 2009
3,120
1,342
[FONT=&quot]Hello everyone. I’m working on the following problem involving a geometric progression and a quadrilateral:[/FONT]
[FONT=&quot]Problem: [/FONT]
[FONT=&quot]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.[/FONT]
[FONT=&quot]Here’s what I’ve done with this problem so far:[/FONT]
[FONT=&quot]Let the lengths of the four sides of the quadrilateral be denoted by t_1 < t_2 < t_3 < t_4.[/FONT]
[FONT=&quot]t_4 = 81 cm (Given)[/FONT]
[FONT=&quot]t_1 + t_2 + t_3 + t_4 = 120 cm (Given Perimeter of Quadrilateral)[/FONT]
[FONT=&quot]Substituting for t_4 above:[/FONT]
[FONT=&quot]t_1 + t_2 + t_3 + 81 = 120[/FONT]
[FONT=&quot]t_1 + t_2 + t_3 = 39[/FONT]
[FONT=&quot]a + ar + ar^2 = 39[/FONT]
[FONT=&quot]a ( 1 + r + r^2 ) = 39[/FONT]
[FONT=&quot]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.[/FONT]
[FONT=&quot]Any and all help will be greatly appreciated. I’m a little rusty at this after many years of keeping away from math.[/FONT]
[FONT=&quot]Thank you.[/FONT]
[FONT=&quot]Shahz.[/FONT]
Alternatively,

for a geometric series,

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

The sum of the first 4 terms is 120

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

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

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

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

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

\(\displaystyle 39r^4-39=120r^3-120\)

\(\displaystyle 120r^3-39r^4=120-39=81\)

\(\displaystyle r^3(120-39r)=81\)

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

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

Hence r=3 is a solution as this gives \(\displaystyle 3^3(40-39)=3^3\)
 
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Reactions: pollardrho06
Feb 2010
36
2
New Jersey, USA
Alternatively,

for a geometric series,

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

The sum of the first 4 terms is 120

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

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

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

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

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

\(\displaystyle 39r^4-39=120r^3-120\)

\(\displaystyle 120r^3-39r^4=120-39=81\)

\(\displaystyle r^3(120-39r)=81\)

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

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

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

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