How Does Radian Measure Relate to Sector Area in a Circle?

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Homework Help Overview

The discussion revolves around the relationship between the radian measure and the area of a sector in a circle, specifically in the context of a circle with a given perimeter. Participants are attempting to prove a specific area formula related to the sector.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between the perimeter of the sector and its area, questioning the validity of the proposed area formula. Some suggest methods to express the angle in terms of the radius, while others raise concerns about the assumptions made regarding the angle and arc length.

Discussion Status

The discussion is ongoing, with various interpretations of the problem being explored. Some participants have offered alternative formulations of the problem, while others have pointed out potential inconsistencies in the original statement. There is no explicit consensus yet on the correct approach or interpretation.

Contextual Notes

Participants note the importance of clearly defining the terms and relationships involved, particularly regarding the angle and arc length, as these are crucial for the proof being sought. There are also mentions of formatting issues in the posts that may affect clarity.

resresd
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in a circle with perimeter of 20cm a sector has radius r, angle θ radians and area Acm^2
prove that A= 25-(r-5)^2

A=1/2 r^2 θ
c=2Πr

the only thing that I can think to do is c=2Πr therefore 20=2Πr
10=Πr r=10/Π
so then A=1/2 r^2 θ
=1/2 (10/Π)^2 θ
= θ (100/2Π^2)
 
Last edited:
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Why don't you post the entire question again, and in a neat way, it will make life easier for everybody here, and surely you will get more responses!

EDIT: you can just edit the first post, and delete the others, or ask a PF to delete the others.
 
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i can't seem to get the 2s to show up superscript...is there a particular way?
 
you might want to use LATEX if you are familiar with that at all? or you can just do it
r^2, and we will know that r is raised to the power of 2.
 
thanks, i hope this is better
 
Honestly, it looks to me like that statement is not true and so cannot be proved. For one thing, it does not involve "[itex]\theta[/itex]" and that can't be right.
 
I believe that the statement should have been like this:

A sector of a circle has a perimeter of 20cm and a area of "A" and the circle has a radius of "r". Prove that "A = 25-(r-5)(r-5)"

Now with simple methods, show that the arc of sector is (20-2r). Find the angle of sector in terms of "r". Then substitute the angle to the area of sector and it is proved easily.
 
If you are not given the angle, how could you possibly prove that the length of the arc of the sector is 20- 2r? The length obviously depends on [itex]\theta[/itex].
 
Indeed it does, but the perimeter also allows to find the Arc length, since the perimeter is the addition of the three sides (Arc length and two radii):

(Length of Arc) + (2 * radius) = Perimeter of Sector

or, Length of Arc = Perimeter of Sector - (2 * radius)

or, Length of Arc = 20 - 2r
 
  • #10
OH! The original post read "a circle with perimeter of 20cm" and when you changed it to "A sector of a circle has a perimeter of 20cm " I was still thinking about the circled and just thought about the length of the arc.
 

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