Green shaded area=π(R-r)(R+r)... It's only a problem to find R and r. You perfectly solved this:)).Thank you for your nice task, Professor! God bless you and your family💯👍👋
Since there are no variables provided in the problem that the answer should be expressed in terms of, it means that if the problem is correct, then the area of the green circle is a constant. You can then just choose the special case where AD is a diameter of outer circle. Meaning that the green area should be pi * 8^2 - pi * 4^2 = pi * (64 - 16) = 48 pi
If R is the radius of the larger circle, r is the radius of the smaller, and x is the perpendicular distance from O to the line, then by the Pythagorean Theorem x^2 + 4^2 = r^2 and x^2 + 8^2 = R^2 => pi(R^2) - pi(r^2) = 48pi
At first, it seems that the radii of the inner and outer circles, let's designate them as r and R, must be found, from which the inner's circle's area can be deducted from the outer circle's, to determine the green area. However, further examination reveals that there is a range of values for r and R which will satisfy the problem's givens. One of them is the special case of O being on BC. In that case, r = 4 and R = 8, the green area is πR² - πr² = π(8)² - π(4)² = (64)π - (16)π = 48π. If this were a multiple choice test, you could make the assumption the the problem statement implies that the solution for the special case applies to all cases and select 48π as the correct answer. As for proving that 48π is the correct answer for all valid cases, PreMath does excellent work in the video, where he shows that (R² - r²) is constant over the valid range of values for r paired with a value of R, with (R² - r²) having a value of 48. (Actually, for every value of r>=4, there is a corresponding value of R.) As an aside, the spelling of donut was discussed on the TV program "Wheel of Fortune". They spelled it "doughnut" in a puzzle, presumably to make the puzzle more challenging to solve. The original spelling was doughnut, which is still considered a correct and preferred spelling in many dictionaries. It got shortened to donut, which is the spelling I've seen almost universally in the USA.
Starting at 4:30 in the video, the intersecting chords theorem can also be used to find the same results. Extend DE both ways to intersect the edges of the larger circle. Then (R+h) * (R-h) = 8*8. or R^2 - h^2 = 64. Following the pythagorean formula on OEC, h^2 + 4^2 = r^2, or r^2 - h^2 = 16. Subtracting the two equations produces R^2 - r^2 = 48.
Let R is Radius big circle and r is Radius small circle Green area=πR^2-πr^2=π(R^2-r^2) in small circle: Let OE=x (r+x)(r-x)=16 r^2-x^2=16 (1) in big circle (R+x)(R-x)=64 R^2-x^2=64 (2) (2) - (1) R^2-x^2-r^2+x^2=64-16=48 So R^2-r^2=48 So Green area=48π square units=150.80 square units.❤❤❤ Thanks Sir
Con centro en "O" trazamos un círculo interior tangente a la cuerda BC→ Área de la corona verde =(Área de la corona delimitada por el círculo exterior y el nuevo círculo) - (Área de la corona delimitada por el círculo interior de la figura propuesta y el nuevo círculo) =π(AD/2)²-π(BC/2)² =π(8²-4²) =π(64-16)=48π ud². Interesante acertijo. Gracias y un saludo.
I can't understand why is too long the solution for some people. Don't you remember the formula of circular ring area???? respect to the chord ??? A = ¼πc²
1/ Area of the green region= pi(sqrR-sqr) 2/ Build the diameter AOA’ intersecting the small circle at points E and F. We have: AExAF=ABxAC (secant theorem) or (R-r)x(R+r)=4x12=48 -> sqR-sqr=48 -> Area of the green region=48 pi sq units😅
Note that the perpendicular distance from the line to O is not specified. If we assume nonetheless that this is a well-posed problem (that the people creating it didn't mess up), then it must be hte case that the answer does not depend on this distance. So we are free to set it to anything that still allows us to formulate the problem properly - we can choose to simplify the solution. I choose to choose that the line passes THROUGH O. That is, our length 16 line is a diameter of the big circle. So the big circle radius is 8 and the small circle radius is 4. So, the green area is pi*(8^2-4^2) = pi*(64-16) = 48*pi. In our heads we can note that 48*3 is 144, 0.1*48 is 4.8, and 1.5 times 4.8 is 7.2, so the area will be a bit less than 151. The actual answer turns out to be 150.796. See how much you can do in your head if you know your way around numbers? And you can do this, in this case, in well under half a minute - it probably took me between 15 and 20 seconds (took a lot longer to type this description). This "unspecified parameter" trick is a really valuable one. It's more useful in "test / quiz" environments, because in some real-world problem you come up with yourself you might not be able to be sure that you can get an answer without knowing that parameter. But in a quiz if they don't GIVE IT TO YOU, then you can generally assume the answer doesn't depend on it. And that can allow for POWERFUL simplification.
Just a suggestion... What about expanding this equation to show how to find the other unknowns after the area was found? I am always trying to find all unknowns, if possible, because I am very rusty, for professional engineering/programming uses and thought it could be a bonus? I really enjoy the equations.
...if those are the only given values and we know there is a unique answer, the distance from the center is irrelevant, so you can just do (8sq - 4sq) Pi
Let's find the area: . .. ... .... ..... Let M be the midpoint of AD. In this case OM is perpendicular to AD and BC as well. Therefore M is also the midpoint of BC and the triangles AMO/DMO and BMO/CMO are right triangles. With R and r being the radii of the outer and inner circle, respectively, we obtain by applying the Pythagorean theorem: AD = AB + BC + CD = 4 + 8 + 4 = 16 AM = AD/2 = 16/2 = 8 BM = BC/2 = 8/2 = 4 R² = AO² = AM² + MO² ∧ r² = BO² = BM² + MO² ⇒ R² − r² = AM² − BM² = 8² − 4² = 64 − 16 = 48 Now we are able to calculate the area of the green region: A(green) = A(outer circle) − A(inner circle) = πR² − πr² = π(R² − r²) = 48π Best regards from Germany
Isn't it really easier to state: R² - 8² = k² = S² - 4² . ... where R is radius of large circle S is radius of small circle K is distance from center to middle of line AD Then, having that, just rearrange things... (R² - S²) = (8² - 4²) ... = 64 - 16 = 48 Having that ... then turn it back into circles ... Area = pi * radius^2 and the difference is as the video shows RING = πR² - πS² ... = π(R² - S²) ... = π(48) = 48π which is approximately 150.80 And we're done.
@@akosibrothercoolpzzle123 I agree-Euler gave us quite a few things from the base of the natural logarithm to his formula detailing the polar form of complex numbers.
Sir your videos are really interesting 👌🏻👌🏻👌🏻🙏🙏🙏
Glad to hear that!
Thanks for the feedback ❤️🙏
Green shaded area=π(R-r)(R+r)... It's only a problem to find R and r. You perfectly solved this:)).Thank you for your nice task, Professor! God bless you and your family💯👍👋
Since there are no variables provided in the problem that the answer should be expressed in terms of, it means that if the problem is correct, then the area of the green circle is a constant. You can then just choose the special case where AD is a diameter of outer circle.
Meaning that the green area should be pi * 8^2 - pi * 4^2 = pi * (64 - 16) = 48 pi
Excellent!
Thanks for sharing ❤️
I really enjoyed watching you solve this. Thank you!
Glad you enjoyed it!
Thanks for the feedback ❤️
If R is the radius of the larger circle, r is the radius of the smaller, and x is the perpendicular distance from O to the line, then by the Pythagorean Theorem
x^2 + 4^2 = r^2 and
x^2 + 8^2 = R^2
=> pi(R^2) - pi(r^2) = 48pi
Excellent!
Thanks for sharing ❤️
Beautiful question!! 👍
Glad to hear that!
Thanks for the feedback ❤️
At first, it seems that the radii of the inner and outer circles, let's designate them as r and R, must be found, from which the inner's circle's area can be deducted from the outer circle's, to determine the green area. However, further examination reveals that there is a range of values for r and R which will satisfy the problem's givens. One of them is the special case of O being on BC. In that case, r = 4 and R = 8, the green area is πR² - πr² = π(8)² - π(4)² = (64)π - (16)π = 48π. If this were a multiple choice test, you could make the assumption the the problem statement implies that the solution for the special case applies to all cases and select 48π as the correct answer. As for proving that 48π is the correct answer for all valid cases, PreMath does excellent work in the video, where he shows that (R² - r²) is constant over the valid range of values for r paired with a value of R, with (R² - r²) having a value of 48. (Actually, for every value of r>=4, there is a corresponding value of R.)
As an aside, the spelling of donut was discussed on the TV program "Wheel of Fortune". They spelled it "doughnut" in a puzzle, presumably to make the puzzle more challenging to solve. The original spelling was doughnut, which is still considered a correct and preferred spelling in many dictionaries. It got shortened to donut, which is the spelling I've seen almost universally in the USA.
Excellent!
Thanks for the awesome feedback ❤️
Starting at 4:30 in the video, the intersecting chords theorem can also be used to find the same results. Extend DE both ways to intersect the edges of the larger circle. Then (R+h) * (R-h) = 8*8. or R^2 - h^2 = 64. Following the pythagorean formula on OEC, h^2 + 4^2 = r^2, or r^2 - h^2 = 16. Subtracting the two equations produces R^2 - r^2 = 48.
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Thanks for sharing ❤️
Intersecting chords:
draw diameter D through pt C.
Long segment is R + r
Short segment is R - r
(R + r)(R -r) = 4 * 12
R^2 - r^2 = 48
Boom.
Very simple. 👍 Great!
Excellent!
Thanks for sharing ❤️
Sir,you are a great teacher.
woooow
Sir u are a magician 😊
So nice of you, dear🌹
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That’s very good thanks for you Sir
Very good method for solve ❤❤❤❤
You are very welcome!🌹
Thanks for the feedback ❤️
R^2=h^2+8^2...r^2=h^2+4^2...AAgreen=π(R^2-r^2)=π(64-16)=48π
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Thanks for sharing ❤️
Let R is Radius big circle
and r is Radius small circle
Green area=πR^2-πr^2=π(R^2-r^2)
in small circle:
Let OE=x
(r+x)(r-x)=16
r^2-x^2=16 (1)
in big circle
(R+x)(R-x)=64
R^2-x^2=64 (2)
(2) - (1)
R^2-x^2-r^2+x^2=64-16=48
So R^2-r^2=48
So Green area=48π square units=150.80 square units.❤❤❤ Thanks Sir
Excellent!
You are very welcome!
Thanks for sharing ❤️
I did it just like you, very simple.
Excellent!
Thanks for the feedback ❤️
Con centro en "O" trazamos un círculo interior tangente a la cuerda BC→ Área de la corona verde =(Área de la corona delimitada por el círculo exterior y el nuevo círculo) - (Área de la corona delimitada por el círculo interior de la figura propuesta y el nuevo círculo) =π(AD/2)²-π(BC/2)² =π(8²-4²) =π(64-16)=48π ud².
Interesante acertijo. Gracias y un saludo.
Excellent!
Thanks for sharing ❤️
A = ¼πC² - ¼πc² = ¼π (C² - c²)
A = ¼π (16² - 8²)
A = 48π cm² ( Solved √ )
I can't understand why is too long the solution for some people.
Don't you remember the formula of circular ring area???? respect to the chord ???
A = ¼πc²
Excellent!
Thanks for sharing ❤️
Thank you!
You are very welcome!
Thanks ❤️
1/ Area of the green region= pi(sqrR-sqr)
2/ Build the diameter AOA’ intersecting the small circle at points E and F.
We have: AExAF=ABxAC (secant theorem)
or (R-r)x(R+r)=4x12=48
-> sqR-sqr=48
-> Area of the green region=48 pi sq units😅
Excellent!
Thanks for sharing ❤️
I solved the problem by the same method you used. Area = Pi*(R^2 - r^2) = 48*Pi
Excellent!
Thanks for sharing ❤️
Note that the perpendicular distance from the line to O is not specified. If we assume nonetheless that this is a well-posed problem (that the people creating it didn't mess up), then it must be hte case that the answer does not depend on this distance. So we are free to set it to anything that still allows us to formulate the problem properly - we can choose to simplify the solution.
I choose to choose that the line passes THROUGH O. That is, our length 16 line is a diameter of the big circle. So the big circle radius is 8 and the small circle radius is 4. So, the green area is pi*(8^2-4^2) = pi*(64-16) = 48*pi. In our heads we can note that 48*3 is 144, 0.1*48 is 4.8, and 1.5 times 4.8 is 7.2, so the area will be a bit less than 151. The actual answer turns out to be 150.796. See how much you can do in your head if you know your way around numbers? And you can do this, in this case, in well under half a minute - it probably took me between 15 and 20 seconds (took a lot longer to type this description).
This "unspecified parameter" trick is a really valuable one. It's more useful in "test / quiz" environments, because in some real-world problem you come up with yourself you might not be able to be sure that you can get an answer without knowing that parameter. But in a quiz if they don't GIVE IT TO YOU, then you can generally assume the answer doesn't depend on it. And that can allow for POWERFUL simplification.
Why to specify, it's clear we can draw a perpendicular to AD from O . Bro it's basic math 🤨
Green Shaded Region = π.R² - π.r² = π (R² - r²) ...¹
By The Chords Theorem
(R + r) (R - r) = 12 . 4
R² - r² = 48
Green Shaded Region = π (R² - r²) = 48π Square Units
Excellent!
Thanks for sharing ❤️
Just a suggestion... What about expanding this equation to show how to find the other unknowns after the area was found? I am always trying to find all unknowns, if possible, because I am very rusty, for professional engineering/programming uses and thought it could be a bonus? I really enjoy the equations.
Great question!
Thanks for the feedback ❤️
...if those are the only given values and we know there is a unique answer, the distance from the center is irrelevant, so you can just do (8sq - 4sq) Pi
Thanks for the feedback ❤️
I solved it. I feel like a genius.
Bravo!
Glad to hear that!
Thanks for the feedback ❤️
R=sqrt(73), r=5, h=3
If it can be set h=0, it could be easier to solve with R=8, r=4.
Thanks for the feedback ❤️
Let's find the area:
.
..
...
....
.....
Let M be the midpoint of AD. In this case OM is perpendicular to AD and BC as well. Therefore M is also the midpoint of BC and the triangles AMO/DMO and BMO/CMO are right triangles. With R and r being the radii of the outer and inner circle, respectively, we obtain by applying the Pythagorean theorem:
AD = AB + BC + CD = 4 + 8 + 4 = 16
AM = AD/2 = 16/2 = 8
BM = BC/2 = 8/2 = 4
R² = AO² = AM² + MO²
∧ r² = BO² = BM² + MO²
⇒ R² − r² = AM² − BM² = 8² − 4² = 64 − 16 = 48
Now we are able to calculate the area of the green region:
A(green) = A(outer circle) − A(inner circle) = πR² − πr² = π(R² − r²) = 48π
Best regards from Germany
Super job!👍
Thanks for sharing ❤️
Isn't it really easier to state:
R² - 8² = k² = S² - 4² . ... where
R is radius of large circle
S is radius of small circle
K is distance from center to middle of line AD
Then, having that, just rearrange things...
(R² - S²) = (8² - 4²) ... = 64 - 16 = 48
Having that ... then turn it back into circles ... Area = pi * radius^2 and the difference is as the video shows
RING = πR² - πS² ... = π(R² - S²) ... = π(48) = 48π which is approximately 150.80
And we're done.
48π square units , nice problem ❤
48 pi. Sq. Units
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Thanks for sharing ❤️
Doughnut is the correct spelling 😊
Pythagoras was the greatest mathematical genius
One of*
@@akosibrothercoolpzzle123 I agree-Euler gave us quite a few things from the base of the natural logarithm to his formula detailing the polar form of complex numbers.
Very true!
Thanks for the feedback ❤️
I thought arc BC was 8, not line BC. That messed me up.