#### emmaiskool242

##### Junior Member

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A=18,C=30,B=?

Can anyone Help Me??

Can anyone Help Me??

- Thread starter emmaiskool242
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- Mar 29, 2006

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A=18,C=30,B=?

Can anyone Help Me??

Can anyone Help Me??

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The Pythagorean Theorem, "a<sup>2</sup> + b<sup>2</sup> = c<sup>2</sup>", is fairly plug-n-chug. So just plug in the values they gave you, and chug away to the answer. Where are you stuck?

Please reply showing all of your steps so far. Thank you.

Eliz.

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ok so 18 squared=324

and 30 squared=900

so 324+b(squared)=900

so you would minus the 324 from both sides

so 900-324=576

so then you would square 576

and then your answer would equal 24

RIGHT?

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a2 + b2 = c2 is the formula correct??

You already have your answer but I thought you might be interested in some Pythagorean Theorem triviaemmaiskool242 said:A=18,C=30,B=?

Can anyone Help Me??

History records that Pythagoras and Diophantus were probably the two most well known mathematicians that had anything to do with right triangles with integer sides. The famous Pythagorean Theorem states that in any right-angled triangle, the sum of the squares of the two legs is equal to the square of the hypotenuse. Another way of stating it is that the area of the square constructed on the long side of a right triangle is equal to the area of the two squares created on the two shorter sides. This is expressed by x^2 + y^2 = z^2.

Almost everyone has heard of the famous 3-4-5 right triangle where 3^2 + 4^2 = 5^2, the simplest, and most fundamental triangle based on the Pythagorean theorem. Surprisingly, fewer people know of the 5, 12, and 13 right triangle, or the 7, 24, and 25 right triangle, which also satisfy the relationship and without any proportional relationship to the 3, 4, and 5 triangle. These sets of three integers, all satisfying the Pythagorean theorem, are traditionally referred to as Pythagorean Triples, of which there are an infinite number.

Primitive Pythagorean Triples of the form x^2 + y^2 = z^2 can be derived from x = m^2 - n^2, y = 2mn, and z = m^2 + n^2 where m and n are arbitrary positive integers of opposite parity (one odd one even), and m is greater than n. (For x, y, & z to be a primitive solution, m and n must have no common factors and must not both be even or odd. Dropping these two limitations will produce non-primitive Pythagorean Triples.

Another series of equations often used for generating triples are x = pq, y = (p^2 - q^2)/2 and z = (p^2 + q^2)/2 where p and q are odd integers with no common factors and p > q > 1.

Consecutive values of m and n will produce triples with one leg and the hypotenuse consecutive. An m and n of 2 and 1 produce the 3-4-5 triple. 3 and 2 produce the 5-12-13 triple. 4 and 3 produce the 7-24-25 triple. Make a table of more and see if you can discover the pattern of the results.

Using the x and y values from the consecutive m and n triples produce triples where the hypotenuse is a perfect square. An m and n of 4 and 3 produces the 7-24-25 triple (25 = 5^2). An m and n of 12 and 5 produces the 119-120-169 triple (169 = 13^2).

Using the y and z values from the consecutive m and n triples produce triples with the smallest leg a perfect square. An m and n of 5 and 4 produces the 9-40-41 triple. An m and n of 13 and 12 produces the 25-312-313 triple.

Consecutive legs can be derived from the m and n values derived from the general expression ni = 2(n(p-1) + m(p-2) and mi = 2m(p-1) + n(p-1). The subscript p indicates the "previous" value. The values m(p-1) and n(p-1) are the previous m and n values while m(p-2) and n(p-2) are the values previous to m(p-1) and n(p-1). Consider the starting point of the 3-4-5 triple which derives from m and n values of 2 and 1. The next triple that will produce a triple with both legs consecutive comes from the values of n = 2(1) + 0 = 2 and m = 2(2) + 1 = 5 which produces the 21-20-29 triple. The next will derive from n = 2(2) + 1 = 5 and m = 2(5) + 2 = 12 producing the 119-120-169 triple. See how many you can create and see if you can determine another interesting relationship between the results.

* Other unique Pythagorean Triples can also be derived from m and n values based on the triangular numbers T = n(n+1)/2, i.e., 1, 3, 6, 10, 15, 21, etc. Using consecutive Triangular numbers for m and n, the triples that result have the smallest leg a perfect cube.

* The product of the two legs of a right triangle is equal to the product of the hypotenuse and the altitude to the hypotenuse.

* Triples can also be derived from x = 2n + 1, y = 2n^2 + 2n, and z = 2n^2 + 2n + 1 where n is any integer. The formulas creat only triangles where the hypotenuse exceeds the larger leg by one.

* Another set of expressions that produce triples is x = n^2, y = (n^2 - 1)^2/2, and z = (n^2 + 1)^2/2.

* Given any integer one can find another such that the sum of their squares is an integral square. Given the number A is odd - Resolve A into any 2 of its factors, equating the larger to m + n and the smaller to M - n. Solve for m and n and substitute into the formulas of 1) above. Given the number A is even - equate it to 2mn and let m and n be any two integers which will result in the product being equal to A/2.

* Fermat discovered that Pythagorean Triples can be derived by expressing the odd squares as the sum of consecutive integers. Using any odd square, express it as x + (x + 1) = N = n^2 and n^2 + x^2 = (x + 1)^2. Another way of looking at it is to say a^2 = (b + c), b and c being consecutive integers. Since (b - c) = 1, we can write a^2 = (b + c)(b - c) = b^2 - c^2 from which a^2 + b^2 = c^2. For example, 25 = 12 + 13 = 5^2 and 5^2 + 12^2 = 13^2 or 121 = 60 + 61 = 11^2 and 11^2 + 60^2 = 61^2.

* Others can be derived by taking any consecutive odd or even numbers and adding their reciprocals as with x and y, we get 1/x + 1/y = (x + y)/xy. It follows that (x + y)^2 + (xy)^2 = N^2. Example, from 5 and 7 we get 1/5 + 1/7 = 12/35 and 12^2 + 35^2 = 37^2 or from 6 and 8 we get 1/6 + 1/8 = 7/24 and 7^2 + 24^2 = 25^2.

* In every triplet of integers for the sides of a Primitive Pythagorean triangle, one of them is always divisible by 3 and one by 5. Also, the product of the two legs is always divisible by 12, and the product of all three sides is always divisible by 60.