Area of a carousel platform [closed]
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I heard about a problem of finding the area of a carousel platform. A tape could not be stretched across the middle because of the machinery but a tape just tangent to the inner circle formed a cord of the outer circle that measured 70'. The answer said, "Suppose the inner circle had a diameter of zero. Then the cord would, in fact, be a diameter so $pi*35^2$ is the area," and it works no matter how large the inner circle. I don't understand how the logic can be extended beyond an inner circle of no diameter. How does this work?
geometry
asked Dec 31 '18 at 19:18
poetasispoetasis
363217
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closed as unclear what you're asking by amWhy, Adrian Keister, Lord Shark the Unknown, mrtaurho, José Carlos Santos Jan 1 at 15:39
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
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I heard about a problem of finding the area of a carousel platform. A tape could not be stretched across the middle because of the machinery but a tape just tangent to the inner circle formed a cord of the outer circle that measured 70'. The answer said, "Suppose the inner circle had a diameter of zero. Then the cord would, in fact, be a diameter so $pi*35^2$ is the area," and it works no matter how large the inner circle. I don't understand how the logic can be extended beyond an inner circle of no diameter. How does this work?
geometry
asked Dec 31 '18 at 19:18
poetasispoetasis
363217
$endgroup$
closed as unclear what you're asking by amWhy, Adrian Keister, Lord Shark the Unknown, mrtaurho, José Carlos Santos Jan 1 at 15:39
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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I first heard of this problem in the early 2000's, when it was presented on an episode of the "Car Talk" series run by the Magliozzi brothers on NPR (National Public Radio) out of Boston. They set it as a problem posed to a summer worker at an amusement park who was charged by the boss with painting the deck of the carousel. Wishing to impress the boss, and saving money by not buying too much paint, the worker came up with the solution outlined by Andrei below. Incidentally (and irrelevantly), I heard this episode while riding in my car just a few miles from the Kingston-Poughkeepsie bridge.
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– Chris Leary
Dec 31 '18 at 19:46
2
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This is reminiscent of the "sneaky" solution to the hole-in-the sphere puzzle
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– saulspatz
Dec 31 '18 at 21:38
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My question was answered to my satisfaction. Why is it now put on hold by amWhy, Adrian Keister, Lord Shark the Unknown, mrtaurho, José Carlos Santos ?
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– poetasis
Jan 1 at 16:54
add a comment |
$begingroup$
I heard about a problem of finding the area of a carousel platform. A tape could not be stretched across the middle because of the machinery but a tape just tangent to the inner circle formed a cord of the outer circle that measured 70'. The answer said, "Suppose the inner circle had a diameter of zero. Then the cord would, in fact, be a diameter so $pi*35^2$ is the area," and it works no matter how large the inner circle. I don't understand how the logic can be extended beyond an inner circle of no diameter. How does this work?
geometry
asked Dec 31 '18 at 19:18
poetasispoetasis
363217
$endgroup$
I heard about a problem of finding the area of a carousel platform. A tape could not be stretched across the middle because of the machinery but a tape just tangent to the inner circle formed a cord of the outer circle that measured 70'. The answer said, "Suppose the inner circle had a diameter of zero. Then the cord would, in fact, be a diameter so $pi*35^2$ is the area," and it works no matter how large the inner circle. I don't understand how the logic can be extended beyond an inner circle of no diameter. How does this work?
geometry
geometry
asked Dec 31 '18 at 19:18
poetasispoetasis
363217
asked Dec 31 '18 at 19:18
poetasispoetasis
363217
asked Dec 31 '18 at 19:18
poetasispoetasis
363217
asked Dec 31 '18 at 19:18
poetasispoetasis
363217
asked Dec 31 '18 at 19:18
poetasispoetasis
363217
363217
closed as unclear what you're asking by amWhy, Adrian Keister, Lord Shark the Unknown, mrtaurho, José Carlos Santos Jan 1 at 15:39
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by amWhy, Adrian Keister, Lord Shark the Unknown, mrtaurho, José Carlos Santos Jan 1 at 15:39
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
I first heard of this problem in the early 2000's, when it was presented on an episode of the "Car Talk" series run by the Magliozzi brothers on NPR (National Public Radio) out of Boston. They set it as a problem posed to a summer worker at an amusement park who was charged by the boss with painting the deck of the carousel. Wishing to impress the boss, and saving money by not buying too much paint, the worker came up with the solution outlined by Andrei below. Incidentally (and irrelevantly), I heard this episode while riding in my car just a few miles from the Kingston-Poughkeepsie bridge.
$endgroup$
– Chris Leary
Dec 31 '18 at 19:46
2
$begingroup$
This is reminiscent of the "sneaky" solution to the hole-in-the sphere puzzle
$endgroup$
– saulspatz
Dec 31 '18 at 21:38
$begingroup$
My question was answered to my satisfaction. Why is it now put on hold by amWhy, Adrian Keister, Lord Shark the Unknown, mrtaurho, José Carlos Santos ?
$endgroup$
– poetasis
Jan 1 at 16:54
add a comment |
$begingroup$
I first heard of this problem in the early 2000's, when it was presented on an episode of the "Car Talk" series run by the Magliozzi brothers on NPR (National Public Radio) out of Boston. They set it as a problem posed to a summer worker at an amusement park who was charged by the boss with painting the deck of the carousel. Wishing to impress the boss, and saving money by not buying too much paint, the worker came up with the solution outlined by Andrei below. Incidentally (and irrelevantly), I heard this episode while riding in my car just a few miles from the Kingston-Poughkeepsie bridge.
$endgroup$
– Chris Leary
Dec 31 '18 at 19:46
2
$begingroup$
This is reminiscent of the "sneaky" solution to the hole-in-the sphere puzzle
$endgroup$
– saulspatz
Dec 31 '18 at 21:38
$begingroup$
My question was answered to my satisfaction. Why is it now put on hold by amWhy, Adrian Keister, Lord Shark the Unknown, mrtaurho, José Carlos Santos ?
$endgroup$
– poetasis
Jan 1 at 16:54
$begingroup$
I first heard of this problem in the early 2000's, when it was presented on an episode of the "Car Talk" series run by the Magliozzi brothers on NPR (National Public Radio) out of Boston. They set it as a problem posed to a summer worker at an amusement park who was charged by the boss with painting the deck of the carousel. Wishing to impress the boss, and saving money by not buying too much paint, the worker came up with the solution outlined by Andrei below. Incidentally (and irrelevantly), I heard this episode while riding in my car just a few miles from the Kingston-Poughkeepsie bridge.
$endgroup$
– Chris Leary
Dec 31 '18 at 19:46
$begingroup$
I first heard of this problem in the early 2000's, when it was presented on an episode of the "Car Talk" series run by the Magliozzi brothers on NPR (National Public Radio) out of Boston. They set it as a problem posed to a summer worker at an amusement park who was charged by the boss with painting the deck of the carousel. Wishing to impress the boss, and saving money by not buying too much paint, the worker came up with the solution outlined by Andrei below. Incidentally (and irrelevantly), I heard this episode while riding in my car just a few miles from the Kingston-Poughkeepsie bridge.
$endgroup$
– Chris Leary
Dec 31 '18 at 19:46
2
2
$begingroup$
This is reminiscent of the "sneaky" solution to the hole-in-the sphere puzzle
$endgroup$
– saulspatz
Dec 31 '18 at 21:38
$begingroup$
This is reminiscent of the "sneaky" solution to the hole-in-the sphere puzzle
$endgroup$
– saulspatz
Dec 31 '18 at 21:38
$begingroup$
My question was answered to my satisfaction. Why is it now put on hold by amWhy, Adrian Keister, Lord Shark the Unknown, mrtaurho, José Carlos Santos ?
$endgroup$
– poetasis
Jan 1 at 16:54
$begingroup$
My question was answered to my satisfaction. Why is it now put on hold by amWhy, Adrian Keister, Lord Shark the Unknown, mrtaurho, José Carlos Santos ?
$endgroup$
– poetasis
Jan 1 at 16:54
add a comment |
1 Answer
1
active
oldest
votes
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The carousel has an inner radius $r$ and an outer radius $R$. Since the tape is tangent to the inner circle, you form a right angle triangle from the center of the inner circle, to the tangent point om the inner circle, to the intersection with the outer circle. Using Pythagoras' theorem, the length of half of the tape is $$frac l2=sqrt{R^2-r^2}$$
The area of the carousel is $$A=pi R^2-pi r^2=pi(R^2-r^2)=pileft(frac l2right)^2$$
answered Dec 31 '18 at 19:34
AndreiAndrei
12.4k21128
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Thanks, It's so simple, once you know the answer. Thanks again.
$endgroup$
– poetasis
Dec 31 '18 at 19:38
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No problem. It always helps if you draw a picture first.
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– Andrei
Dec 31 '18 at 19:39
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The carousel has an inner radius $r$ and an outer radius $R$. Since the tape is tangent to the inner circle, you form a right angle triangle from the center of the inner circle, to the tangent point om the inner circle, to the intersection with the outer circle. Using Pythagoras' theorem, the length of half of the tape is $$frac l2=sqrt{R^2-r^2}$$
The area of the carousel is $$A=pi R^2-pi r^2=pi(R^2-r^2)=pileft(frac l2right)^2$$
answered Dec 31 '18 at 19:34
AndreiAndrei
12.4k21128
$endgroup$
$begingroup$
Thanks, It's so simple, once you know the answer. Thanks again.
$endgroup$
– poetasis
Dec 31 '18 at 19:38
$begingroup$
No problem. It always helps if you draw a picture first.
$endgroup$
– Andrei
Dec 31 '18 at 19:39
add a comment |
$begingroup$
The carousel has an inner radius $r$ and an outer radius $R$. Since the tape is tangent to the inner circle, you form a right angle triangle from the center of the inner circle, to the tangent point om the inner circle, to the intersection with the outer circle. Using Pythagoras' theorem, the length of half of the tape is $$frac l2=sqrt{R^2-r^2}$$
The area of the carousel is $$A=pi R^2-pi r^2=pi(R^2-r^2)=pileft(frac l2right)^2$$
answered Dec 31 '18 at 19:34
AndreiAndrei
12.4k21128
$endgroup$
$begingroup$
Thanks, It's so simple, once you know the answer. Thanks again.
$endgroup$
– poetasis
Dec 31 '18 at 19:38
$begingroup$
No problem. It always helps if you draw a picture first.
$endgroup$
– Andrei
Dec 31 '18 at 19:39
add a comment |
$begingroup$
The carousel has an inner radius $r$ and an outer radius $R$. Since the tape is tangent to the inner circle, you form a right angle triangle from the center of the inner circle, to the tangent point om the inner circle, to the intersection with the outer circle. Using Pythagoras' theorem, the length of half of the tape is $$frac l2=sqrt{R^2-r^2}$$
The area of the carousel is $$A=pi R^2-pi r^2=pi(R^2-r^2)=pileft(frac l2right)^2$$
answered Dec 31 '18 at 19:34
AndreiAndrei
12.4k21128
$endgroup$
The carousel has an inner radius $r$ and an outer radius $R$. Since the tape is tangent to the inner circle, you form a right angle triangle from the center of the inner circle, to the tangent point om the inner circle, to the intersection with the outer circle. Using Pythagoras' theorem, the length of half of the tape is $$frac l2=sqrt{R^2-r^2}$$
The area of the carousel is $$A=pi R^2-pi r^2=pi(R^2-r^2)=pileft(frac l2right)^2$$
answered Dec 31 '18 at 19:34
AndreiAndrei
12.4k21128
answered Dec 31 '18 at 19:34
AndreiAndrei
12.4k21128
answered Dec 31 '18 at 19:34
AndreiAndrei
12.4k21128
answered Dec 31 '18 at 19:34
AndreiAndrei
12.4k21128
12.4k21128
$begingroup$
Thanks, It's so simple, once you know the answer. Thanks again.
$endgroup$
– poetasis
Dec 31 '18 at 19:38
$begingroup$
No problem. It always helps if you draw a picture first.
$endgroup$
– Andrei
Dec 31 '18 at 19:39
add a comment |
$begingroup$
Thanks, It's so simple, once you know the answer. Thanks again.
$endgroup$
– poetasis
Dec 31 '18 at 19:38
$begingroup$
No problem. It always helps if you draw a picture first.
$endgroup$
– Andrei
Dec 31 '18 at 19:39
$begingroup$
Thanks, It's so simple, once you know the answer. Thanks again.
$endgroup$
– poetasis
Dec 31 '18 at 19:38
$begingroup$
Thanks, It's so simple, once you know the answer. Thanks again.
$endgroup$
– poetasis
Dec 31 '18 at 19:38
$begingroup$
No problem. It always helps if you draw a picture first.
$endgroup$
– Andrei
Dec 31 '18 at 19:39
$begingroup$
No problem. It always helps if you draw a picture first.
$endgroup$
– Andrei
Dec 31 '18 at 19:39
add a comment |
$begingroup$
I first heard of this problem in the early 2000's, when it was presented on an episode of the "Car Talk" series run by the Magliozzi brothers on NPR (National Public Radio) out of Boston. They set it as a problem posed to a summer worker at an amusement park who was charged by the boss with painting the deck of the carousel. Wishing to impress the boss, and saving money by not buying too much paint, the worker came up with the solution outlined by Andrei below. Incidentally (and irrelevantly), I heard this episode while riding in my car just a few miles from the Kingston-Poughkeepsie bridge.
$endgroup$
– Chris Leary
Dec 31 '18 at 19:46
2
$begingroup$
This is reminiscent of the "sneaky" solution to the hole-in-the sphere puzzle
$endgroup$
– saulspatz
Dec 31 '18 at 21:38
$begingroup$
My question was answered to my satisfaction. Why is it now put on hold by amWhy, Adrian Keister, Lord Shark the Unknown, mrtaurho, José Carlos Santos ?
$endgroup$
– poetasis
Jan 1 at 16:54