Calculating Perimeter: Circle Sidewalk with a Radius of 220m and Width of 30m

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Homework Help Overview

The problem involves calculating the perimeter of a circular sidewalk surrounding a roundabout with a specified radius and width. The original poster presents a scenario where the radius is 220 meters and the sidewalk width is 30 meters, leading to confusion regarding the correct interpretation of the dimensions and the resulting perimeter.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore different interpretations of the dimensions, questioning whether the given radius is accurate or if it should be considered as a diameter. There are attempts to apply the perimeter formula for circles and rectangles, with some participants suggesting the need to clarify the shape of the sidewalk.

Discussion Status

The discussion is ongoing, with various interpretations being explored. Some participants have offered insights into potential misunderstandings regarding the dimensions, while others have raised questions about the nature of the sidewalk (straight versus circular). There is no clear consensus yet on how to proceed with the calculations.

Contextual Notes

Participants note the ambiguity in the problem statement regarding whether the sidewalk is circular or straight, and there is a mention of the need to consider both the inner and outer diameters when calculating the perimeter. The original poster has indicated that they have not yet resolved the problem.

Femme_physics
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Homework Statement



http://img94.imageshack.us/img94/9216/thethingy.jpg

Through a circle roundabout whose radius is 220 meters, they constructed a sidewalk in width of 30 meters. Calculate the perimeter of the sidewalk.


The Attempt at a Solution



perimeter of a circle equation: 2 x pi x R

So 220+30 = 250

2 x pi x 250 = 1570.8 squared meters.


And yet I'm told the answer is 880 meters.
 
Last edited by a moderator:
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Hi Femme_physics! :smile:

I think the sidewalk must be straight across the middle. :redface:

(and don't forget it has two sides :wink:)
 
Hi Fp! :smile:

Is it possible that the roundabout has a diameter of 220 m, instead of a radius of 220 m? :confused:
 
Hi tiny-timmy! :)

How confusing -- since they drew another circle that says "sidewalk" so I thought the sidewalk is round.

So it really should be

(250+30) x 2 = 560

Hmm...still off the mark
 
Hi ILS! :)
I like Serena said:
Hi Fp! :smile:

Is it possible that the roundabout has a diameter of 220 m, instead of a radius of 220 m? :confused:
You're right! I was lost in translation!

If I presume the sidewalk is round it should be

2 x pi x (250/2) = 785.398

Argh no still off
 
Do you have half a sidewalk around the roundabout? :confused:

Isn't that a bit narrow to walk on?

A couple of mistakes wouldn't be able to walk next to each other!
 
Last edited:
30 meters width is a bit narrow?

And why half? That formula relates the entire perimeter based on radius
 
Femme_physics said:
30 meters width is a bit narrow?

And why half? That formula relates the entire perimeter based on radius

So what is the proper radius?

First of the roundabout and then with the sidewalk included?
 
Hey Femme, the question is a little vague in this regard: is the sidewalk straight or circular?

If the side walk is straight:
2*(220*x)+2*(30*y) = 880
I don't have all the math figured out (hence the variables), but just work with me here.

The perimeter of a rectangular object is:
2*l+2*w
No big deal. We know that.

We also know that an object with a length of 440m (our diameter), and no width (but 2 sides, somehow), has a perimeter of 880m.

But, when you place a rectangle of any width over top of a circle's center, you have 4 "ears" hanging off of the edge which are dead weight in this simulation.
If the sidewalk is straight, I'm willing to wager that the changed length of the 440m diameter being offset left and right by 15 meters each is balanced by the arc-length of the two ends of the sidewalk in order to still equal 880m, since only the width is 30m, not the arc-lengths.

However, if the sidewalk is circular and 30 meters wide and equals 880m in perimeter, then we can just write an equation to solve for its radii.

880 = 2r_{1}\pi + 2(r_{1}+30)\pi

Unless I was careless, this works out to:

r_{1} \approx 55m

So, we've got to be getting some misinformation here.
 
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  • #10
Hi ILS! :smile:
I like Serena said:
Is it possible that the roundabout has a diameter of 220 m, instead of a radius of 220 m? :confused:

Yup, it's the diameter …

(220 + 60) * 22/7 :biggrin:​
 
  • #11
tiny-tim said:
Hi ILS! :smile:


Yup, it's the diameter …

(220 + 60) * 22/7 :biggrin:​

Don't you need 2 circumferences to determine the perimeter of a "donut" shape?
 
  • #12
Hi AJKing! :smile:
AJKing said:
Don't you need 2 circumferences to determine the perimeter of a "donut" shape?

Yup! :biggrin:

Rubbish question, isn't it? :rolleyes:​
 
  • #13
tiny-tim said:
Hi AJKing! :smile:


Yup! :biggrin:

Rubbish question, isn't it? :rolleyes:​

Lol, quite so!
 
  • #14
AJKing said:
Hey Femme, the question is a little vague in this regard: is the sidewalk straight or circular?

If the side walk is straight:
2*(220*x)+2*(30*y) = 880
I don't have all the math figured out (hence the variables), but just work with me here.

The perimeter of a rectangular object is:
2*l+2*w
No big deal. We know that.

We also know that an object with a length of 440m (our diameter), and no width (but 2 sides, somehow), has a perimeter of 880m.

But, when you place a rectangle of any width over top of a circle's center, you have 4 "ears" hanging off of the edge which are dead weight in this simulation.
If the sidewalk is straight, I'm willing to wager that the changed length of the 440m diameter being offset left and right by 15 meters each is balanced by the arc-length of the two ends of the sidewalk in order to still equal 880m, since only the width is 30m, not the arc-lengths.

However, if the sidewalk is circular and 30 meters wide and equals 880m in perimeter, then we can just write an equation to solve for its radii.

880 = 2r_{1}\pi + 2(r_{1}+30)\pi

Unless I was careless, this works out to:

r_{1} \approx 55m

So, we've got to be getting some misinformation here.

The fact that the perimeter equals 880 is not something I know -- it's just from the solution manual, that's what I need to find
 
  • #15
Femme_physics said:
The fact that the perimeter equals 880 is not something I know -- it's just from the solution manual, that's what I need to find

You get that perimeter if you calculate with 220 m as the diameter of the inner circle. So what is the diameter of the outer circle?

ehild
 
  • #16
If you were going to provide walkers with a short-cut over a circular patch of precious grass, you would provide 2 intersecting straight paths, one running N-S and the other E-W. (you get the picture, since it's an intersection of some description)

If you then wanted to border that walking area with a distinctive coloured paving stone, you would need to know that perimeter, i.e., the sum of the lengths of all 12 sides. I make this 8.8 x 102 metres. o:)
 
  • #17
Hey fp! :smile:

Did you already solve this problem?
Were you able to work through the posts of the different people and work out what the solution is?
 
  • #18
Hi ILS :)

I didn't, but we already had the math test 2 days ago and now I got some other priorities.

So thanks everyone for your replies and the great help you were giving! I just skipped a similar question that was on the test and took a similar one to the "diamond" shape that I also posted here at about the same time. That was easier :)

Thanks for caring!
 
  • #19
tiny-tim said:
Yup, it's the diameter ...
(220 + 60) * 22/7

I believe tiny-tim & ILS have already made a reasonable guess at the author's intent.
Summarizing: Around a circular roundabout whose diameter is 220 meters,
is constructed a circular sidewalk of width 30 meters.
Calculate the outer perimeter of the sidewalk.

The diameter of the inner-circle is 220m
the width of the sidewalk is given = 30m.
The perimeter = 2 \pi r = 2 \pi (\frac {220}{2}+30).
(or if you prefer perimeter = \pi d) as tiny-tim indicated above).
 

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