Calculating the Area of a Circle Segment with Given Radius and Angle

  • Thread starter Thread starter skybox
  • Start date Start date
  • Tags Tags
    Area Circle
Click For Summary

Homework Help Overview

The discussion revolves around calculating the area of a circle segment given a radius of 7 and an angle of 150°. Participants are exploring the relationships between the area of the circle, the sector, and the triangle formed within the segment.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the formulas for the area of a circle, sector, and triangle, with some attempting to apply the 30-60-90 triangle rule. Others question the validity of this approach and suggest using the SAS formula for triangles instead.

Discussion Status

There is an ongoing exploration of the correct formulas and methods to find the area of the segment. Some participants have provided guidance on drawing diagrams to clarify the relationships between the areas involved, while others have pointed out errors in the original calculations.

Contextual Notes

Participants are working within the constraints of homework rules, which may limit the information they can use or the methods they can apply. There is also a noted confusion regarding the assumptions about the triangle's properties and the area calculations.

skybox
Messages
37
Reaction score
0
Hi all, I am having an issue trying to solve the following problem

Homework Statement


PkM0RFH.png

I know that the radius of the circle is 7 and the angle of the segment is 150°



Homework Equations


Area of a circle: A = \pi{r}^2
Area of the sector of the circle: A = \frac{n}{360}\pi r^{2}
Area of a triangle: A = \frac{n}{360}\pi r^{2}


The Attempt at a Solution


I know to get the final solution to get the area of that particular part of the circle, I need to subtract the whole area of the circle with the segment of the circle (the triangular portion).

To get the area of the triangle portion, I subtract the area of the sector of the circle with the area of the triangle of that portion multiplied by 2 (since there are essentially two triangles). Also, to get the area of the triangular portion, I can use the 30-60-90 rule. To get this:
Short leg:
A_s = \frac{1}{2}H = \frac{1}{2}*7 = \frac{7}{2}
Long leg:
A_l=\frac{1}{2}H\sqrt{3} = \frac{7}{2}\sqrt{3}
Area of whole triangle:
A_t=\frac{1}{2}*\frac{7}{2}*\frac{7}{2}\sqrt{3}*2=\frac{49}{4}\sqrt{3}

Now the area of the whole sector is:
A_t = \frac{150}{360}49\pi-\frac{49}{4}\sqrt{3}

To get the area of just that portion of the circle the question is asking, subtract the area of the whole circle with the answer from above:
A = 49\pi-(\frac{150}{360}49\pi-\frac{49}{4}\sqrt{3})

I was wondering if I am doing this problem correctly? I am not getting any of the answers in the possible solutions. Thanks!
 
Physics news on Phys.org
skybox said:
Hi all, I am having an issue trying to solve the following problem

Homework Statement


PkM0RFH.png

I know that the radius of the circle is 7 and the angle of the segment is 150°



Homework Equations


Area of a circle: A = \pi{r}^2
Area of the sector of the circle: A = \frac{n}{360}\pi r^{2}
Area of a triangle: A = \frac{n}{360}\pi r^{2}


The Attempt at a Solution


I know to get the final solution to get the area of that particular part of the circle, I need to subtract the whole area of the circle with the segment of the circle (the triangular portion).

To get the area of the triangle portion, I subtract the area of the sector of the circle with the area of the triangle of that portion multiplied by 2 (since there are essentially two triangles). Also, to get the area of the triangular portion, I can use the 30-60-90 rule. To get this:
Short leg:
A_s = \frac{1}{2}H = \frac{1}{2}*7 = \frac{7}{2}

But that half triangle is not a 30-60-90 triangle.
 
LCKurtz said:
But that half triangle is not a 30-60-90 triangle.

Hmmm, yes you are right. Looks like I made a mistake. Can I assume it is a right triangle if I cut it in half?
 
skybox said:
Hmmm, yes you are right. Looks like I made a mistake. Can I assume it is a right triangle if I cut it in half?

You are given that one leg of the triangle is a radius, that is enough to be sure the triangle is isosceles.
 
  • Like
Likes   Reactions: 1 person
skybox said:
Hmmm, yes you are right. Looks like I made a mistake. Can I assume it is a right triangle if I cut it in half?

Yes, but why would you want to deal with angles like ##15^\circ##? What is the formula for area of a triangle given two sides and the included angle ##\theta##? Draw a picture and figure it out if you don't know the formula. It's what you need here.
 
  • Like
Likes   Reactions: 1 person
LCKurtz said:
Yes, but why would you want to deal with angles like ##15^\circ##? What is the formula for area of a triangle given two sides and the included angle ##\theta##? Draw a picture and figure it out if you don't know the formula. It's what you need here.

Thanks. I researched this and it looks like I can use the SAS (Side Angle Side) formula to get the area of this triangle. That formula is:
A=\frac{1}{2}ab\sin C
Which, for my problem, is equal to:
A_t = \frac{1}{2}(7)(7)\sin 150\degree
A_t = \frac{1}{2}(49)(0.5)
A_t = \frac{49}{4}

Now to get the area of that small portion of the circle, I take the area of the whole circle and subtract it with the area of the triangle:
A = 49\pi - \frac{49}{4}
A = 49(\pi - \frac{1}{4})

Does this look right because this answer does not equal any of the possible answers for the problem. Thanks again!
 
In your expression A=49π−49/4, you have taken the area of the whole circle, and not a fraction of it.
 
  • Like
Likes   Reactions: 1 person
pongo38 said:
In your expression A=49π−49/4, you have taken the area of the whole circle, and not a fraction of it.

Thanks for the reply pongo38. I might be confusing myself. If I take area of the whole circle and subtract it with the area of the triangle, won't that give me the area of the smaller circle?
 
I think it would help if you drew the whole circle, then the 150 degree segment, then the triangle. Put on the diagram all the areas you can identify.
 
  • Like
Likes   Reactions: 1 person
  • #10
pongo38 said:
I think it would help if you drew the whole circle, then the 150 degree segment, then the triangle. Put on the diagram all the areas you can identify.

Thanks pongo. I drew it out and understand why my thinking was incorrect on trying to subtract the whole circle.

So I drew it out:
LHtI8nc.png


Now I am confused on how to continue to approach this problem. I am not sure if the the portion of (B) is exactly a half circle, which I cannot assume given the problem. I even tried to work it out and didn't get the correct problem with the same thinking. Any suggestions? Thanks
 
  • #11
Go back to your original post. In the Relevant equations section, you have stated the correct formula for the area of a circular sector. The formula for the area of the triangle is incorrect.

From the diagram, the area of the shaded portion = the area of the circular sector - the area of the triangle.

You have calculated the area of the circular sector as (150/360)*49 * pi
You have calculated the area of the triangle as 49/4

Can you calculate the area of the shaded portion now? Remember to simplify fractions in your final result.
 
  • Like
Likes   Reactions: 1 person
  • #12
skybox said:
Thanks pongo. I drew it out and understand why my thinking was incorrect on trying to subtract the whole circle.

So I drew it out:
LHtI8nc.png


Now I am confused on how to continue to approach this problem. I am not sure if the the portion of (B) is exactly a half circle, which I cannot assume given the problem. I even tried to work it out and didn't get the correct problem with the same thinking. Any suggestions? Thanks

Area B has nothing to do with the problem. Find the area of the sector of the circle bounded by the two radii and subtract the area of the triangle from it.
 
  • Like
Likes   Reactions: 1 person
  • #13
Thanks guys! I finally solved it and got a better understanding of the problem.

Attached is the solution I created in word when trying to solve this problem. Thanks again for everyone's help.
 

Attachments

  • Capture.jpg
    Capture.jpg
    11.2 KB · Views: 523

Similar threads

  • · Replies 24 ·
Replies
24
Views
3K
  • · Replies 10 ·
Replies
10
Views
25K
Replies
30
Views
5K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 12 ·
Replies
12
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
23
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K