Sides of a Quadrilateral

**pollardrho06** · May 20th, 2010, 10:42 AM

Hello everyone. I’m 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.
Here’s what I’ve 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. I’m a little rusty at this after many years of keeping away from math.
Thank you.
Shahz.

**bjhopper** · May 20th, 2010, 01:14 PM

Originally Posted by pollardrho06

Hello everyone. I’m 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.
Here’s what I’ve 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. I’m 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

**pollardrho06** · May 20th, 2010, 01:41 PM

Originally Posted by bjhopper

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?

**Archie Meade** · May 20th, 2010, 01:47 PM

Originally Posted by pollardrho06

Hello everyone. I’m 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.
Here’s what I’ve 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. I’m a little rusty at this after many years of keeping away from math.
Thank you.
Shahz.

.

**bjhopper** · May 20th, 2010, 05:58 PM

[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

**pollardrho06** · May 21st, 2010, 02:56 AM

[quote=bjhopper;516333]

Originally Posted by pollardrho06

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.

**Archie Meade** · May 21st, 2010, 04:18 AM

Originally Posted by pollardrho06

Hello everyone. I’m 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.
Here’s what I’ve 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. I’m 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 $3^3(40-39)=3^3$

**pollardrho06** · May 21st, 2010, 04:39 AM

Originally Posted by Archie Meade

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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