POTD
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Puzzle of the day. A fence of a house is 10m length (square) At point d a dog is tied, it can't go inside the house, when it moves along the sides it reaches the point x (max), Where the point x, and d are a midpoints of there sides. Q : What is the area that the dog can move in ?
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LongHC wrote:
Q : What is the area that the dog can move in ?
What's the area of the house (the actual building within the fence)? Without knowing that, you cannot calculate the area the dog can walk in. Oops - sorry. I thought the dog was tied inside the fence.
Regards, Nish
Nish’s thoughts on MFC, C++/CLI and .NET (my blog)
My latest book : C++/CLI in Action / Amazon.com link -
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:laugh::laugh::laugh: Now you are giving Quartz a taste of his own medicine. BTW: Is the answer: 95.66
Rama Krishna Vavilala wrote:
BTW: Is the answer: 95.66
Wrong :->
Rama Krishna Vavilala wrote:
Now you are giving Quartz a taste of his own medicine.
It seems not only he will taste it.:-O
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LongHC wrote:
Q : What is the area that the dog can move in ?
What's the area of the house (the actual building within the fence)? Without knowing that, you cannot calculate the area the dog can walk in. Oops - sorry. I thought the dog was tied inside the fence.
Regards, Nish
Nish’s thoughts on MFC, C++/CLI and .NET (my blog)
My latest book : C++/CLI in Action / Amazon.com linkNishant Sivakumar wrote:
What's the area of the house (the actual building within the fence)?
It is written in the puzzle.
Nishant Sivakumar wrote:
Without knowing that, you cannot calculate the area the dog can walk in.
Remember it can't go through it, it will only stick around the wall.
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Puzzle of the day. A fence of a house is 10m length (square) At point d a dog is tied, it can't go inside the house, when it moves along the sides it reaches the point x (max), Where the point x, and d are a midpoints of there sides. Q : What is the area that the dog can move in ?
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The length of the leash is 20 m. So the area is the sum of 3 half circles of radius 20m, 15m and 5m. Think about how the leash wraps around the house as the god walks around it.
Using the GridView is like trying to explain to someone else how to move a third person's hands in order to tie your shoelaces for you. -Chris Maunder
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Puzzle of the day. A fence of a house is 10m length (square) At point d a dog is tied, it can't go inside the house, when it moves along the sides it reaches the point x (max), Where the point x, and d are a midpoints of there sides. Q : What is the area that the dog can move in ?
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Total area dog can walk in = (Pi * 20 * 20)/2 + (Pi * 15 * 15)/2 + (Pi * 5 * 5)/2 = Pi/2 * (400 + 225 + 25) = Pi * 325 = approx 1020.5
Regards, Nish
Nish’s thoughts on MFC, C++/CLI and .NET (my blog)
My latest book : C++/CLI in Action / Amazon.com link -
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Nishant Sivakumar wrote:
What's the area of the house (the actual building within the fence)?
It is written in the puzzle.
Nishant Sivakumar wrote:
Without knowing that, you cannot calculate the area the dog can walk in.
Remember it can't go through it, it will only stick around the wall.
LongHC wrote:
Remember it can't go through it, it will only stick around the wall.
Sorry, initially I thought the dog was inside the fence.
Regards, Nish
Nish’s thoughts on MFC, C++/CLI and .NET (my blog)
My latest book : C++/CLI in Action / Amazon.com link -
Rama Krishna Vavilala wrote:
BTW: Is the answer: 95.66
Wrong :->
Rama Krishna Vavilala wrote:
Now you are giving Quartz a taste of his own medicine.
It seems not only he will taste it.:-O
Assuming that the dog cannot cross the fence. The dog will walk in an area like the following: dog.png (1.1 Kb) The area = 2 * area of rectangles + Area of the 1/6 the circle = 2 * 0.5 * 5 * sqrt(75) + 1/6 * pi * 100 Where did I go wrong?
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The length of the leash is 20 m. So the area is the sum of 3 half circles of radius 20m, 15m and 5m. Think about how the leash wraps around the house as the god walks around it.
Using the GridView is like trying to explain to someone else how to move a third person's hands in order to tie your shoelaces for you. -Chris Maunder
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The length of the leash is 20 m. So the area is the sum of 3 half circles of radius 20m, 15m and 5m. Think about how the leash wraps around the house as the god walks around it.
Using the GridView is like trying to explain to someone else how to move a third person's hands in order to tie your shoelaces for you. -Chris Maunder
Andy Brummer wrote:
Think about how the leash wraps around the house as the god walks around it.
SOAPBOX!!! :rolleyes:
Regards, Nish
Nish’s thoughts on MFC, C++/CLI and .NET (my blog)
My latest book : C++/CLI in Action / Amazon.com link -
Puzzle of the day. A fence of a house is 10m length (square) At point d a dog is tied, it can't go inside the house, when it moves along the sides it reaches the point x (max), Where the point x, and d are a midpoints of there sides. Q : What is the area that the dog can move in ?
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1139.271 1021.42771 Forgot to /2 for pi*5*5
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Puzzle of the day. A fence of a house is 10m length (square) At point d a dog is tied, it can't go inside the house, when it moves along the sides it reaches the point x (max), Where the point x, and d are a midpoints of there sides. Q : What is the area that the dog can move in ?
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:laugh: Taking your word, "The most obvious solutions are not right". Also I put this puzzle specially for you.
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oh i am flattered :) i should have given more time to the question btw good one
Omit Needless Words - Strunk, William, Jr.
Vista? Cryptography Next Generation (CNG) here
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LongHC wrote:
Andy Brummer and Nishant Sivakumar got it right. it's 325*PIm
:jig:
Regards, Nish
Nish’s thoughts on MFC, C++/CLI and .NET (my blog)
My latest book : C++/CLI in Action / Amazon.com link -
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Assuming that the dog cannot cross the fence. The dog will walk in an area like the following: dog.png (1.1 Kb) The area = 2 * area of rectangles + Area of the 1/6 the circle = 2 * 0.5 * 5 * sqrt(75) + 1/6 * pi * 100 Where did I go wrong?
Rama Krishna Vavilala wrote:
The dog will walk in an area like the following:
It is not like this, in fact it will move in a circles (Quarters) with radii 20, 15, and 5, then adding them and multiplying by 2 (Because it can go from the 2 sides) will give the solution, which is 325*PIm as Andy and Nish said.
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1139.271 1021.42771 Forgot to /2 for pi*5*5
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Rama Krishna Vavilala wrote:
The dog will walk in an area like the following:
It is not like this, in fact it will move in a circles (Quarters) with radii 20, 15, and 5, then adding them and multiplying by 2 (Because it can go from the 2 sides) will give the solution, which is 325*PIm as Andy and Nish said.
Oh Ok! I also thought that the dog was inside the fence.:doh:
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LongHC wrote:
A fence of a house is 10m length (square)
If 10m is the length of the fence, the side of the square is only 2.5m. PI * 325 is ridiculously large. Or did you mean one side of the square is 10m?
Cheers, Vıkram.
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byalmightybob -
Avery very nice one. I like these things, please come up now with one like it if you have:cool::rose:
