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Posted: 10/30/2010 1:07:42 PM EDT
The perimeter of a rectangle is 25 cm greater than the sum of the length and the width. If the length is 5 cm greater than the width, what are the dimensions of the rectangle?

I guess I've been in the sun too long. What would be the formula to solve this?
Link Posted: 10/30/2010 1:10:41 PM EDT
[#1]
Quoted:
The perimeter of a rectangle is 25 cm greater than the sum of the length and the width. If the length is 5 cm greater than the width, what are the dimensions of the rectangle?

I guess I've been in the sun too long. What would be the formula to solve this?



perimeter = 2L + 2W

. Write everything else out in algebraic form. Solve for L and W
Link Posted: 10/30/2010 1:12:42 PM EDT
[#2]
87


(sorry, couldn't help myself)
Link Posted: 10/30/2010 1:15:38 PM EDT
[#3]

Quoted:






The perimeter of a rectangle is 25 cm greater than the sum of the length and the width. If the length is 5 cm greater than the width, what are the dimensions of the rectangle?
I guess I've been in the sun too long. What would be the formula to solve this?






Units in centimeters.
'P'erimeter = 25 + 'L'ength + 'W'idth
First: P = 25 + L + W




Second: L = 5 + W





Knowing: P = L + W + L + W




Then: L + W = 25
Or just looked at the question and said: "Oh, length and width are 25? That means it must be 10 by 15".
 
Link Posted: 10/30/2010 1:17:58 PM EDT
[#4]
2L + 2W = 25+ (L+W)



L=W+5




2(W+5)+2W=25+(W+5) +W




2W+10+2W=25+2W+5




4W+10=30+2W




2W=20




W=10




2L+20=25+L+10




L=15




W is width

L is length
Link Posted: 10/30/2010 1:18:28 PM EDT
[#5]
Quoted:
The perimeter of a rectangle is 25 cm greater than the sum of the length and the width. If the length is 5 cm greater than the width, what are the dimensions of the rectangle?

I guess I've been in the sun too long. What would be the formula to solve this?


P=2l+2w+25
l=5+w
W=W Do the math from there.
Link Posted: 10/30/2010 1:21:10 PM EDT
[#6]





Quoted:





Quoted:


The perimeter of a rectangle is 25 cm greater than the sum of the length and the width. If the length is 5 cm greater than the width, what are the dimensions of the rectangle?





I guess I've been in the sun too long. What would be the formula to solve this?






P=2l+2w+25


l=5+w


W=W Do the math from there.
Really? Come on now...
 
Link Posted: 10/30/2010 1:25:43 PM EDT
[#7]
2x + 2y = 25 + x + y
Therefore
x + y = 25
...
y - 5 = x
..

Therefore
y -5 + y = 25

Therefore
2y = 30

Therefore
y = 15

Therefore

x = 10

Link Posted: 10/30/2010 1:27:19 PM EDT
[#8]



Quoted:


2x + 2y = 25 + x + y

Therefore

x + y = 25

...

y - 5 = x

..



Therefore

y -5 + y = 25



Therefore

2y = 30



Therefore

y = 15



Therefore



x = 10



Yours is more elegant than mine.






 
Link Posted: 10/30/2010 1:28:27 PM EDT
[#9]
Thanks guys! I was trying to help my ninth grader and I had figured out 15cm and 10cm but couldn't figure out how to write the steps to determine it.

Link Posted: 10/30/2010 1:28:32 PM EDT
[#10]



Quoted:





Quoted:


Quoted:

The perimeter of a rectangle is 25 cm greater than the sum of the length and the width. If the length is 5 cm greater than the width, what are the dimensions of the rectangle?



I guess I've been in the sun too long. What would be the formula to solve this?




P=2l+2w+25

l=5+w

W=W Do the math from there.
Really? Come on now...



 


I believe he was saying to use substitution.



 
Link Posted: 10/30/2010 1:29:26 PM EDT
[#11]
P=25+2W+2(W+5)
Link Posted: 10/30/2010 1:31:27 PM EDT
[#12]



Quoted:


Quoted:


Quoted:


Quoted:

The perimeter of a rectangle is 25 cm greater than the sum of the length and the width. If the length is 5 cm greater than the width, what are the dimensions of the rectangle?



I guess I've been in the sun too long. What would be the formula to solve this?
P=2l+2w+25

l=5+w

W=W Do the math from there.
Really? Come on now...

 
I believe he was saying to use substitution.

 
You mean like substituting W for W? God, that's just brilliant. Or did you mean his first erroneous equation?





 
Link Posted: 10/30/2010 1:31:33 PM EDT
[#13]
Quoted:
2L + 2W = 25+ (L+W)

L=W+5

2(W+5)+2W=25+(W+5) +W

2W+10+2W=25+2W+5

4W+10=30+2W

2W=20

W=10

2L+20=25+L+10

L=15

W is width
L is length


Correct.
Link Posted: 10/30/2010 1:34:35 PM EDT
[#14]
Quoted:
P=25+2W+2(W+5)


That's was the line I was working.

P=25+2W+2(W+5)
P=25+2W+2W+10
P=35+4w
P-35=35+4W-35
P-35=4W (which really doesn't work)
Link Posted: 10/30/2010 1:36:23 PM EDT
[#15]



Quoted:


P=25+2W+2(W+5)
You too??



O.M.G.



The first sentence of the problem:

The perimeter of a rectangle is 25 cm greater than the sum of the length and the width

=

Rectangle Perimeter is 25 greater than sum of Length and Width

=

Perimeter is 25 added to sum of Length and Width

=

Perimeter is 25 added to Length added to Width

=

Perimeter = 25 + Length + Width

=

P = 25 + L + W







 
Link Posted: 10/30/2010 1:36:33 PM EDT
[#16]
Quoted:
2L + 2W = 25+ (L+W)

L=W+5

2(W+5)+2W=25+(W+5) +W

2W+10+2W=25+2W+5

4W+10=30+2W

2W=20

W=10

2L+20=25+L+10

L=15

W is width
L is length


This is how I backed into the answer after my first attempt but I didn't keep the equations in order enough to duplicate it and be able to show the work.
Link Posted: 10/30/2010 1:55:28 PM EDT
[#17]
Quoted:

Quoted:

Quoted:
Quoted:
The perimeter of a rectangle is 25 cm greater than the sum of the length and the width. If the length is 5 cm greater than the width, what are the dimensions of the rectangle?

I guess I've been in the sun too long. What would be the formula to solve this?


P=2l+2w+25
l=5+w
W=W Do the math from there.
Really? Come on now...

 

I believe he was saying to use substitution.
 


It was how I was taught to solve this problems. The book suggested deliniate x for w, but I use w to keep it simple.
Link Posted: 10/30/2010 1:56:04 PM EDT
[#18]
Quoted:

Quoted:
Quoted:
Quoted:
Quoted:
The perimeter of a rectangle is 25 cm greater than the sum of the length and the width. If the length is 5 cm greater than the width, what are the dimensions of the rectangle?

I guess I've been in the sun too long. What would be the formula to solve this?
P=2l+2w+25
l=5+w
W=W Do the math from there.
Really? Come on now...
 
I believe he was saying to use substitution.
 
You mean like substituting W for W? God, that's just brilliant. Or did you mean his first erroneous equation?

 


I was establishing the numbers he needed then he does the distributive property and gets his answer.
Link Posted: 10/30/2010 2:08:27 PM EDT
[#19]



Quoted:



Quoted:


Quoted:


Quoted:


Quoted:


Quoted:

The perimeter of a rectangle is 25 cm greater than the sum of the length and the width. If the length is 5 cm greater than the width, what are the dimensions of the rectangle?



I guess I've been in the sun too long. What would be the formula to solve this?
P=2l+2w+25

l=5+w

W=W Do the math from there.
Really? Come on now...

 
I believe he was saying to use substitution.

 
You mean like substituting W for W? God, that's just brilliant. Or did you mean his first erroneous equation?

 
I was establishing the numbers he needed then he does the distributive property and gets his answer.


The first part in red is wrong.





 
Link Posted: 10/30/2010 2:12:34 PM EDT
[#20]



Quoted:



Quoted:


Quoted:


Quoted:


Quoted:

The perimeter of a rectangle is 25 cm greater than the sum of the length and the width. If the length is 5 cm greater than the width, what are the dimensions of the rectangle?



I guess I've been in the sun too long. What would be the formula to solve this?
P=2l+2w+25

l=5+w

W=W Do the math from there.
Really? Come on now...

 
I believe he was saying to use substitution.

 
It was how I was taught to solve this problems. The book suggested deliniate x for w, but I use w to keep it simple.
So... W = Width? W=W means nothing. Blue = Blue. F = F. Schizophrenic = Schizophrenic. x = x.

"Terms: P = perimeter; W = width; L = length"

followed by the equations, then the written out solution.



If that is what you meant, I can understand that. Maybe you just have a funny way of saying that.





 
Link Posted: 10/30/2010 2:12:36 PM EDT
[#21]
You just gave me a headache.
Link Posted: 10/30/2010 2:17:20 PM EDT
[#22]

Quoted:





The perimeter of a rectangle is 25 cm greater than the sum of the length and the width. If the length is 5 cm greater than the width, what are the dimensions of the rectangle?
I guess I've been in the sun too long. What would be the formula to solve this?





P = Perimeter




L = Length




W = Width
A rectangle has 4 sides: 2 length sides and 2 width sides. The perimeter of a rectangle is the sum of all sides, or (abbreviated) 2L + 2W.
So far, we've arrived at:
1.) P = 2L + 2W
The text says that the perimeter of this rectangle is 25cm greater than the sum of it's length and width.  Another way of phrasing this is "The perimeter is the length, plus the width, plus another 25". Or:
2.) P = L + W + 25
Now that we have a decent definition of what P is supposed to be, let's substitute it in formula #1 for calculating the perimeter of our hypothetical rectangle:
3.) L + W + 25 = 2L + 2W
Hey look, now we're down to two variables: L and W. Sure would be nice if we could just get this down to one variable. Let's keep reading.




"...the length is 5cm greater than the width"
Sweet, I like more info. So, what they're saying is:
4.) L = W + 5
Awesome, now l know what L is supposed to be. Let's substitute L in formula #3, shall we?
W + 5 + W + 25 = 2(W + 5) + 2W
And let's simplify that expression:
2W + 30 = 2(W + 5) + 2W
Sweet, now all we have to do is solve for W.
W + 15 = W + 5 + W




5 + W = 15




W = 10
Awesome. Now that we know the width is 10, we just need to find the length. You might recall in formula #4, that L = W + 5. Well...
L = 10 + 5




L = 15
Therefore, the rectangle dimensions are 15cm long, and 10cm wide. How do I know this is right? Well, let's check our work.









"The perimeter of a rectangle is 25 cm greater than the sum of the length and the width
"
What's the sum of the length and width? Well, it's 25 (L=15 and W=10, remember?).
Is this 25cm greater than the perimeter?  Well....
P = 2W + 2L




P = 2(10) + 2(15)




P = 50
Is 50 25cm greater than the sum of the length and width, 25? I do believe it is.
 
Link Posted: 10/30/2010 2:27:08 PM EDT
[#23]
Quoted:

Quoted:
The perimeter of a rectangle is 25 cm greater than the sum of the length and the width. If the length is 5 cm greater than the width, what are the dimensions of the rectangle?

I guess I've been in the sun too long. What would be the formula to solve this?

P = Perimeter
L = Length
W = Width

A rectangle has 4 sides: 2 length sides and 2 width sides. The perimeter of a rectangle is the sum of all sides, or (abbreviated) 2L + 2W.

So far, we've arrived at:

1.) P = 2L + 2W

The text says that the perimeter of this rectangle is 25cm greater than the sum of it's length and width.  Another way of phrasing this is "The perimeter is the length, plus the width, plus another 25". Or:

2.) P = L + W + 25

Now that we have a decent definition of what P is supposed to be, let's substitute it in formula #1 for calculating the perimeter of our hypothetical rectangle:

3.) L + W + 25 = 2L + 2W

Hey look, now we're down to two variables: L and W. Sure would be nice if we could just get this down to one variable. Let's keep reading.
"...the length is 5cm greater than the width"

Sweet, I like more info. So, what they're saying is:

4.) L = W + 5

Awesome, now l know what L is supposed to be. Let's substitute L in formula #3, shall we?

(W + 5) + W + 25 = 2(W + 5) + 2W

And let's simplify that expression:

2W + 30 = 2(W + 5) + 2W

Sweet, now all we have to do is solve for W.

W + 15 = W + 5 + W
5 + W = 15
W = 10

Awesome. Now that we know the width is 10, we just need to find the length. You might recall in formula #4, that L = W + 5. Well...

L = 10 + 5
L = 15

Therefore, the rectangle dimensions are 15cm long, and 10cm wide. How do I know this is right? Well, let's check our work.

"The perimeter of a rectangle is 25 cm greater than the sum of the length and the width
"

What's the sum of the length and width? Well, it's 25 (L=15 and W=10, remember?).

Is this 25cm greater than the perimeter?  Well....

P = 2W + 2L
P = 2(10) + 2(15)
P = 50

Is 50 25cm greater than the sum of the length and width, 25? I do believe it is.

 

Q. E. D.  Motherfuckers!
Link Posted: 10/30/2010 2:28:42 PM EDT
[#24]
in my head in about 3 seconds, sounds like 10 & 15
Link Posted: 10/30/2010 2:32:01 PM EDT
[#25]



Quoted:

Q. E. D.  Motherfuckers!



Damn right. I'm full of shit until it's demonstrated.



 
Link Posted: 10/30/2010 2:34:15 PM EDT
[#26]
Just a running tally


Quoted: P=2l+2w+25




Quoted: P=25+2W+2(W+5)



Quoted: That's was the line I was working.






 



 

 
Link Posted: 10/30/2010 2:37:07 PM EDT
[#27]
As an engineer I am fapping hard to this thread!
Link Posted: 10/30/2010 2:59:34 PM EDT
[#28]



Quoted:


You just gave me a headache.








How do I reech deese keeeeeds?!!!



 
Link Posted: 10/30/2010 3:03:04 PM EDT
[#29]
Quoted:
The perimeter of a rectangle is 25 cm greater than the sum of the length and the width. If the length is 5 cm greater than the width, what are the dimensions of the rectangle?

I guess I've been in the sun too long. What would be the formula to solve this?


2x + 2y - 25 = x + y

x - 5 = y

ETA:

x = Length
y = Width
Link Posted: 10/30/2010 3:08:47 PM EDT
[#30]







Quoted:




As an engineer I am fapping hard to this thread!




I took the AP Calc exam my Senior year in HS (scored a 4), and stopped there. I never did go to college (well, I started but quit shortly thereafter), and thus, all personal math development ended right then and there. My brain is certainly wired for it, but...use it or lose it, as they say.
The sad thing (it's been 14 years since I've sat in a math class), I doubt I could pass a Math 90 final if you handed one to me right now. I've forgotten a LOT of details.
 
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