Help me!
- Thread starter cllynn213
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The area of a circle segment is given by \(\displaystyle \frac{1}{2}r^{2}({\theta}-sin({\theta}))\).
The formula for the chord length is \(\displaystyle 2rsin(\frac{{\theta}}{2})\)
Set the chord formula equal to 20 and solve for \(\displaystyle {\theta}\)
Then sub \(\displaystyle {\theta}\) into the circle segment formula.
The formula for the chord length is \(\displaystyle 2rsin(\frac{{\theta}}{2})\)
Set the chord formula equal to 20 and solve for \(\displaystyle {\theta}\)
Then sub \(\displaystyle {\theta}\) into the circle segment formula.
pka
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I too am completely stumped!
Why would you be given a problem that requires trigonometry if you have not studied it?
There is a formula for a circular region if one knows the radius and central angle.
However, given the radius and chord length we have to use trigonometry to find the angle. Are you given the central angle?
The area of the triangular region is the area of an isosceles triangle.
The altitude is \(\displaystyle \sqrt {12^2 - 10^2 }\), the 10 is half the base.
Thus the area of the triangular region is \(\displaystyle \left( {\sqrt {12^2 - 10^2 } } \right)\left( 10 \right)\).
Therefore the area of the complete circular sector an somewhat more that that.
But that is an approximate area of a segment of a circle with a radius of 12 meters and the length of the chord is 20 meters.
Why would you be given a problem that requires trigonometry if you have not studied it?
There is a formula for a circular region if one knows the radius and central angle.
However, given the radius and chord length we have to use trigonometry to find the angle. Are you given the central angle?
The area of the triangular region is the area of an isosceles triangle.
The altitude is \(\displaystyle \sqrt {12^2 - 10^2 }\), the 10 is half the base.
Thus the area of the triangular region is \(\displaystyle \left( {\sqrt {12^2 - 10^2 } } \right)\left( 10 \right)\).
Therefore the area of the complete circular sector an somewhat more that that.
But that is an approximate area of a segment of a circle with a radius of 12 meters and the length of the chord is 20 meters.
First, set the chord length formula equal to 20 and solve for \(\displaystyle {\theta}\)
You know the radius, 12 meters.
\(\displaystyle 2(12)sin(\frac{{\theta}}{2})=20\)
\(\displaystyle 24sin(\frac{{\theta}}{2})=20\)
\(\displaystyle sin(\frac{{\theta}}{2})=\frac{5}{6}\)
\(\displaystyle \frac{{\theta}}{2}=sin^{-1}(\frac{5}{6})\)
\(\displaystyle {\theta}=2sin^{-1}(\frac{5}{6})=1.97022156668\)
\(\displaystyle =\frac{29475{\pi}}{46999}\)
Now, put this into the circle segment formula:
You know the radius, 12 meters.
\(\displaystyle 2(12)sin(\frac{{\theta}}{2})=20\)
\(\displaystyle 24sin(\frac{{\theta}}{2})=20\)
\(\displaystyle sin(\frac{{\theta}}{2})=\frac{5}{6}\)
\(\displaystyle \frac{{\theta}}{2}=sin^{-1}(\frac{5}{6})\)
\(\displaystyle {\theta}=2sin^{-1}(\frac{5}{6})=1.97022156668\)
\(\displaystyle =\frac{29475{\pi}}{46999}\)
Now, put this into the circle segment formula:
pka
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I have rethought this whole question!
In this diagram the total ‘colored’ area is the area of the sector.
The area of the two triangles, blue & yellow, is (0.5)(c)(r) where c is the length of the cord. This will give you a much better approximation of the area of the sector.
You are missing the green area.
In this diagram the total ‘colored’ area is the area of the sector.
The area of the two triangles, blue & yellow, is (0.5)(c)(r) where c is the length of the cord. This will give you a much better approximation of the area of the sector.
You are missing the green area.
I'm sorry, I thought you needed the segment, which is what I have, sloppily, shaded in the diagram. Do you need the sector?.
The area for a segment is given by the area of a sector, \(\displaystyle \frac{1}{2}r^{2}{\theta}\), minus the area of an isosceles triangle, \(\displaystyle \frac{1}{2}r^{2}sin({\theta})\). Which gives \(\displaystyle \frac{1}{2}r^{2}{\theta}-\frac{1}{2}r^{2}sin({\theta})=\frac{1}{2}r^{2}({\theta}-sin({\theta}))\).
The area for a segment is given by the area of a sector, \(\displaystyle \frac{1}{2}r^{2}{\theta}\), minus the area of an isosceles triangle, \(\displaystyle \frac{1}{2}r^{2}sin({\theta})\). Which gives \(\displaystyle \frac{1}{2}r^{2}{\theta}-\frac{1}{2}r^{2}sin({\theta})=\frac{1}{2}r^{2}({\theta}-sin({\theta}))\).
Ok lynn... if you are expected to do this without using trig to find the angle at
center in galactus' diagram, then you are probably expected to at least draw
the diagram to scale and measure that angle yourself.
If you do that, you'll get close to 113 degrees; doing this properly using trig
will give you 112.8853~ degrees.
This means the area of the whole thing (colored section in pka's diagram)
is 113/360 times the area of the circle (a circle has total of 360 degrees).
Area of the circle = pi * r^2 = pi * 12^2 = pi * 144 = 3.14 * 144 = 452~
(using 3.14 for pi, since we're approximating)
So area of colored section = 113/360 * 452 = 142~
However, that area includes the area of the 12-12-20 isosceles triangle;
the area of that triangle is 66~ (if you can't calculate that, you're in trouble!).
So that leaves the area of the segment = 142~ - 66~ = 76~
Hope that helps you understand HOW this works.
If trig is used and pi is used accurately, you'd get this:
colored area = 141.856...
triangle area = 66.332...
segment area= 75.524...
center in galactus' diagram, then you are probably expected to at least draw
the diagram to scale and measure that angle yourself.
If you do that, you'll get close to 113 degrees; doing this properly using trig
will give you 112.8853~ degrees.
This means the area of the whole thing (colored section in pka's diagram)
is 113/360 times the area of the circle (a circle has total of 360 degrees).
Area of the circle = pi * r^2 = pi * 12^2 = pi * 144 = 3.14 * 144 = 452~
(using 3.14 for pi, since we're approximating)
So area of colored section = 113/360 * 452 = 142~
However, that area includes the area of the 12-12-20 isosceles triangle;
the area of that triangle is 66~ (if you can't calculate that, you're in trouble!).
So that leaves the area of the segment = 142~ - 66~ = 76~
Hope that helps you understand HOW this works.
If trig is used and pi is used accurately, you'd get this:
colored area = 141.856...
triangle area = 66.332...
segment area= 75.524...
Hey pka, did you use MathCad to post your diagram?. I need something like that, besides paint. I got Geometer's sketchpad, thinking it would be good for this, but I have been unable to get it to post because of a JavaScript issue or something. It turned out to be a nuisance.
Thanks to all that replied to this problem. Your answers were very helpful and the explanations even better.
I'm taking Education Direct (now Penn Foster) courses and the books are not very self explanatory. I have been really disappointed in their course material in electrical engineering. I'll be taking trig right after I get done with these geometry modules and I must say that I am not looking forward to it. I'm certain I'll be posting on here for help with trig.
Once again...THANK YOU!!!!
Chris
I'm taking Education Direct (now Penn Foster) courses and the books are not very self explanatory. I have been really disappointed in their course material in electrical engineering. I'll be taking trig right after I get done with these geometry modules and I must say that I am not looking forward to it. I'm certain I'll be posting on here for help with trig.
Once again...THANK YOU!!!!
Chris
Hey Denis, hope you had a nice Christmas.
\(\displaystyle 2sqrt{11}(10)=66.332\) is the area of an isosceles.
Also, \(\displaystyle \frac{1}{2}(144)sin(1.97022156668)=66.332\) is the area of an isosceles.
This is just another isosceles area formula besides the one you used. I found the derivation in an old trig/alg book. The image is below. That's easier tham typing the LaTex.
First, I solved for theta in the chord formula, which can be seen by my previous post. theta is 1.97022156668 radians or 112.88538 degrees.
The formula for a segment, what I have shaded in my haphazard diagram, is given by \(\displaystyle \frac{1}{2}r^{2}({\theta}-sin({\theta}))\). It just takes the sector area and subtracts the triangle area, giving segment area.
Which gives:\(\displaystyle \frac{1}{2}(144)(1.97022156668-sin(1.97022156668))=75.523456994 m^{2}\)
We can check our answers by finding the middle ordinate, \(\displaystyle r(1-cos(\frac{\theta}{2}))=12(1-cos(.98511078334))=5.36675041931=>12-5.36675041931=2\sqrt{11}\)
Calc:
\(\displaystyle 2\int_{-12}^{-2sqrt{11}}sqrt{144-x^{2}}dx=75.5234569935\)
Sorry, if I got carried away. It's just fun to check different methods to see if it comes out the same. It did.
\(\displaystyle 2sqrt{11}(10)=66.332\) is the area of an isosceles.
Also, \(\displaystyle \frac{1}{2}(144)sin(1.97022156668)=66.332\) is the area of an isosceles.
This is just another isosceles area formula besides the one you used. I found the derivation in an old trig/alg book. The image is below. That's easier tham typing the LaTex.
First, I solved for theta in the chord formula, which can be seen by my previous post. theta is 1.97022156668 radians or 112.88538 degrees.
The formula for a segment, what I have shaded in my haphazard diagram, is given by \(\displaystyle \frac{1}{2}r^{2}({\theta}-sin({\theta}))\). It just takes the sector area and subtracts the triangle area, giving segment area.
Which gives:\(\displaystyle \frac{1}{2}(144)(1.97022156668-sin(1.97022156668))=75.523456994 m^{2}\)
We can check our answers by finding the middle ordinate, \(\displaystyle r(1-cos(\frac{\theta}{2}))=12(1-cos(.98511078334))=5.36675041931=>12-5.36675041931=2\sqrt{11}\)
Calc:
\(\displaystyle 2\int_{-12}^{-2sqrt{11}}sqrt{144-x^{2}}dx=75.5234569935\)
Sorry, if I got carried away. It's just fun to check different methods to see if it comes out the same. It did.
Yikes!
I simply hate the word "theta" and/or its symbol; enough to terrify most students :evil:
LET THE ANGLE FORMED AT CENTER OF CIRCLE BY THE 2 RADIUS LINES
FROM THE CENTER TO THE ENDS OF THE CHORD = G.
Given radius r and chord c:
G = asin(c / 2r)
A = area of isosceles triangle = c*sqrt(4r^2 - c^2) / 4
S = area of sector = pi*r^2*G / 180
Segment area = S - A
= (pi*r^2*G - 180*A) / 180
My view; I don't welcome others :wink:
I simply hate the word "theta" and/or its symbol; enough to terrify most students :evil:
LET THE ANGLE FORMED AT CENTER OF CIRCLE BY THE 2 RADIUS LINES
FROM THE CENTER TO THE ENDS OF THE CHORD = G.
Given radius r and chord c:
G = asin(c / 2r)
A = area of isosceles triangle = c*sqrt(4r^2 - c^2) / 4
S = area of sector = pi*r^2*G / 180
Segment area = S - A
= (pi*r^2*G - 180*A) / 180
My view; I don't welcome others :wink: