Area portion of circle

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


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:
[tex]A_s = \frac{1}{2}H = \frac{1}{2}*7 = \frac{7}{2}[/tex]
Long leg:
[tex]A_l=\frac{1}{2}H\sqrt{3} = \frac{7}{2}\sqrt{3}[/tex]
Area of whole triangle:
[tex]A_t=\frac{1}{2}*\frac{7}{2}*\frac{7}{2}\sqrt{3}*2=\frac{49}{4}\sqrt{3}[/tex]

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

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:
[tex]A = 49\pi-(\frac{150}{360}49\pi-\frac{49}{4}\sqrt{3})[/tex]

I was wondering if I am doing this problem correctly? I am not getting any of the answers in the possible solutions. Thanks!
 

Answers and Replies

  • #2
LCKurtz
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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: [tex]A = \pi{r}^2[/tex]
Area of the sector of the circle: [tex]A = \frac{n}{360}\pi r^{2}[/tex]
Area of a triangle: [tex]A = \frac{n}{360}\pi r^{2}[/tex]


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:
[tex]A_s = \frac{1}{2}H = \frac{1}{2}*7 = \frac{7}{2}[/tex]

But that half triangle is not a 30-60-90 triangle.
 
  • #3
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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?
 
  • #4
verty
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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.
 
  • #5
LCKurtz
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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.
 
  • #6
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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:
[tex]A=\frac{1}{2}ab\sin C[/tex]
Which, for my problem, is equal to:
[tex]A_t = \frac{1}{2}(7)(7)\sin 150\degree [/tex]
[tex]A_t = \frac{1}{2}(49)(0.5)[/tex]
[tex]A_t = \frac{49}{4}[/tex]

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:
[tex]A = 49\pi - \frac{49}{4}[/tex]
[tex]A = 49(\pi - \frac{1}{4})[/tex]

Does this look right because this answer does not equal any of the possible answers for the problem. Thanks again!
 
  • #7
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In your expression A=49π−49/4, you have taken the area of the whole circle, and not a fraction of it.
 
  • #8
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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?
 
  • #9
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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.
 
  • #10
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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
SteamKing
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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.
 
  • #12
LCKurtz
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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.
 
  • #13
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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.
 

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