|
|
|
|
|
|
|
|
|
 Posted: Sat Feb 17, 2007 12:55 pm
The Pythagorean Theorem indicates the A squared plus B squared = C; now this equation can be used to find the slanted side of a right triangle. When conducting this equation, you find A, multiply it by itself, and find B, and multiply it by itself. Then add the two together to give you the sum for C.
But the equation is not concluded yet. After finding C, you must then find the square root of C. One easy way to do this is to list al of the S.R.s on a line:
4 9 16 25 36 49 64 81 100 121 144 169 ..... and so on until you cannot find any more. Then, if C is equal to one of these numbers, you have been lucky and got the easy way out.
If C is not one of the numbers, but comes in between (an imperfect square), you take these steps:
1. Find the factors of C, and make sure one of the factors is a square root of any number. If both of the numbers are square roots, then you can find a number higher than the one you already found. Just always be sure to USE THE HIGHEST FACTOR THAT IS A SQUARE ROOT OF ANY KIND.
2. After you found the highest square root, and a number that isn't the square root, you put the square root sign over the 2 numbers. Then reduce the highest square root to it's factor. (ex. highest number: 49 reduced to: 7)
3. Then remove the square root sign from the reduced number, but leave it on the other number.
You have now found the square root of C, and have completed the Pythagorean Theorym. The value of C is the slanted side of any right triangle, and is always a little bit longer than the other two.
Thank you for reading about the Pythagorean Theorym, if I spelled it right.
|
 |
|
|
|
|
|
|
|
|
|
|
|
|
|
Posted: Sun Feb 18, 2007 12:49 pm
wow you sure do know about pre algebra. I did not know about the Pythagorean Theorem. Thanks for explaining it so clearly and easily to me.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Posted: Tue Feb 20, 2007 1:37 pm
I recently saw how the Pythagorean Theorem was used to prove that length can contract and time can slow down...
Ah relativity.
I never knew that it was so simple.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Posted: Wed Feb 21, 2007 2:19 am
Much simpler way.
Take any right-angled triangle.
Label the hypotenuse(/side opposite the right angle(/longest side)) c.
Label the other two sides a and b.
Solve a²+b² = c²
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Posted: Fri Feb 23, 2007 4:21 pm
Dave the lost Much simpler way. Take any right-angled triangle. Label the hypotenuse(/side opposite the right angle(/longest side)) c. Label the other two sides a and b. Solve a²+b² = c² DUH.... but that was at the beginning of what i mentioned, then you've got to find the square root of c!
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Posted: Sat Feb 24, 2007 6:01 am
Incidentally, √ should be a square root sign. If it's something else or doesn't appear at all, please tell me along with your browser/OS. Doomstyle Dave the lost Much simpler way. Take any right-angled triangle. Label the hypotenuse(/side opposite the right angle(/longest side)) c. Label the other two sides a and b. Solve a²+b² = c² DUH.... but that was at the beginning of what i mentioned, then you've got to find the square root of c! Surely this post is mainly about finding the square root of a number then, rather than usage of the Pythagorean theorem? Well, just my little critique, which some may view as a bit harsh, but, well, for starters your post is really unclear. Your suggestion for finding the root of a number isn't that helpful or clear. Basically, as far as I can tell, your advice is: 1. Write out a long list of the squares. If it's one of them, work out which whole number is the root (with the last bit inferred). 2. If that doesn't work, stick a square root sign over the number and simplify, with a poorly written method of telling people how to simplify, which isn't actually correct. First of all, you use square root all through your procedure, when you should be using square. For example: Quote: 1. Find the factors of C, and make sure one of the factors is a square root of any number. If both of the numbers are square roots, then you can find a number higher than the one you already found. Just always be sure to USE THE HIGHEST FACTOR THAT IS A SQUARE ROOT OF ANY KIND. Any number is the square root of itself squared, so it applies to every number. Secondly, your entire explanation of how to simplify could be written much more clearly as: √ab = √a . √b, thus something like √32 = √16.√2 = 4√2 To make it even more clear, we can tell people that √a²b = a√b Notice how I compressed three paragraphs down into a single line, most of which is an example of which you didn't include any, much to the detriment of readability. Equations are used precisely because they're so much easier to use for many maths related things. Equation: √a²b = a√b Words: For any number given by the square of the product of a number and the square of a number, it can be written as the product of the first number and the square of the second. Do you see how the equation is so much simpler and clearer?
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Posted: Sat Feb 24, 2007 8:28 pm
The fun part is when you have to generalize to non-Euclidean spaces, and the meaning of the Pythagorean theorem changes or becomes invalid.
Consider a sphere, say, all the points in 3-space at unit distance from the origin. Lines on a sphere are the great circles, circles along the sphere that are centered at the origin.
In this case, the Pythagorean theorem doesn't apply, because you can create an equilateral triangle with 3 90-degree interior angles. Just put one vertex at (1, 0, 0), one vertex at (0, 1, 0) and one vertex at (0, 0, 1).
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Posted: Sun Feb 25, 2007 8:09 am
In the non Euclidean stuff you lose pythagoras by gain the beautiful result
Area of triangle with angles A, B and C = A+B+C-π
for the sphere you mentioned and
Area of triangle with angles A, B and C = π-(A+B+C)
for the hyperbolic plane, which is the coolest thing ever, at least in the field of geometry-that-I'm-aware-of.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Posted: Sun Feb 25, 2007 9:58 am
jestingrabbit In the non Euclidean stuff you lose pythagoras by gain the beautiful result Area of triangle with angles A, B and C = A+B+C-π for the sphere you mentioned and Area of triangle with angles A, B and C = π-(A+B+C) for the hyperbolic plane, which is the coolest thing ever, at least in the field of geometry-that-I'm-aware-of. what's n in this case? and how are the lenghts of the sides of the triangle with angles A, B, and C incorporated into the equation for the area? And also is there a pythagorean equivalent for non-euclidian space?
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Posted: Sun Feb 25, 2007 10:57 pm
poweroutage jestingrabbit In the non Euclidean stuff you lose pythagoras by gain the beautiful result Area of triangle with angles A, B and C = A+B+C-π for the sphere you mentioned and Area of triangle with angles A, B and C = π-(A+B+C) for the hyperbolic plane, which is the coolest thing ever, at least in the field of geometry-that-I'm-aware-of. what's n in this case? and how are the lenghts of the sides of the triangle with angles A, B, and C incorporated into the equation for the area? And also is there a pythagorean equivalent for non-euclidian space? I think what you're reading as an n is a π (pi). The lengths are incorporated because the angles fully determine the triangles: similar triangles don't occur in spherical or hyperbolic geometry. I'll have to think a while to come up with the modified pythagorean identity. Edit: If you've got a right angled triangle on a sphere with sides of length θ and φ, then the third side has length arccos(cos(θ)cos(φ)), which you can get by using dot products. It might have a simpler form, but I don't know what it is.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Posted: Tue Feb 27, 2007 7:39 pm
jestingrabbit poweroutage jestingrabbit In the non Euclidean stuff you lose pythagoras by gain the beautiful result Area of triangle with angles A, B and C = A+B+C-π for the sphere you mentioned and Area of triangle with angles A, B and C = π-(A+B+C) for the hyperbolic plane, which is the coolest thing ever, at least in the field of geometry-that-I'm-aware-of. what's n in this case? and how are the lenghts of the sides of the triangle with angles A, B, and C incorporated into the equation for the area? And also is there a pythagorean equivalent for non-euclidian space? I think what you're reading as an n is a π (pi). The lengths are incorporated because the angles fully determine the triangles: similar triangles don't occur in spherical or hyperbolic geometry. I'll have to think a while to come up with the modified pythagorean identity. Edit: If you've got a right angled triangle on a sphere with sides of length θ and φ, then the third side has length arccos(cos(θ)cos(φ)), which you can get by using dot products. It might have a simpler form, but I don't know what it is. you're gonna derive an equivalent pythagorean theorm for non-euclidian geometry? ... awsome! lol .... well if you actually do it, good luck, I'm sure you wont need it though.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Posted: Wed Feb 28, 2007 1:43 pm
Dave the lost Incidentally, √ should be a square root sign. If it's something else or doesn't appear at all, please tell me along with your browser/OS. Doomstyle Dave the lost Much simpler way. Take any right-angled triangle. Label the hypotenuse(/side opposite the right angle(/longest side)) c. Label the other two sides a and b. Solve a²+b² = c² DUH.... but that was at the beginning of what i mentioned, then you've got to find the square root of c! Surely this post is mainly about finding the square root of a number then, rather than usage of the Pythagorean theorem? Well, just my little critique, which some may view as a bit harsh, but, well, for starters your post is really unclear. Your suggestion for finding the root of a number isn't that helpful or clear. Basically, as far as I can tell, your advice is: 1. Write out a long list of the squares. If it's one of them, work out which whole number is the root (with the last bit inferred). 2. If that doesn't work, stick a square root sign over the number and simplify, with a poorly written method of telling people how to simplify, which isn't actually correct. First of all, you use square root all through your procedure, when you should be using square. For example: Quote: 1. Find the factors of C, and make sure one of the factors is a square root of any number. If both of the numbers are square roots, then you can find a number higher than the one you already found. Just always be sure to USE THE HIGHEST FACTOR THAT IS A SQUARE ROOT OF ANY KIND. Any number is the square root of itself squared, so it applies to every number. Secondly, your entire explanation of how to simplify could be written much more clearly as: √ab = √a . √b, thus something like √32 = √16.√2 = 4√2 To make it even more clear, we can tell people that √a²b = a√b Notice how I compressed three paragraphs down into a single line, most of which is an example of which you didn't include any, much to the detriment of readability. Equations are used precisely because they're so much easier to use for many maths related things. Equation: √a²b = a√b Words: For any number given by the square of the product of a number and the square of a number, it can be written as the product of the first number and the square of the second. Do you see how the equation is so much simpler and clearer? not really
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Posted: Mon Mar 05, 2007 8:46 am
Sorry, but I'm with Dave the Lost on this one.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Posted: Mon Apr 16, 2007 1:04 pm
The Pythagorean Theorem. A squared plus B squared equals C squared. But you all must remember the fact that the triangle must be a RIGHT triangle.
The theory is very helpful when it comes to engineering and other jobs that may involve shapes.
To utilize this theorem, you must find a right triangle with a missing side length. Then you do A squared+B squared= C squared.
Let me remind everyone that the square root sign is also called The Radical Sign, Mathematically speaking.
///////////////////////MATH RULES!!!\\\\\\\\\\\\\\\\
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Reply
