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Pythagorean triples are given by the formulas x2 - y2, 2xy, and x2 + y2. Use the formulas for the Pythagorean triples to prove why it is not possible for a right triangle to have legs with lengths of 16 and an odd number. Show all of your work for full credit.

Verified Answer :

LammettHash

Let \(k\) be an integer. Suppose there is a triangle with legs of length 16 and \(2k+1\). Then by the Pythagorean theorem, the length of the hypotenuse should be

\(\sqrt{16^2+(2k+1)^2}=\sqrt{4k^2+4k+257}\)

The formulas for Pythagorean triples say that if the legs are integers, then so must be the hypotenuse, because if \(x=16\) and \(y=2k+1\) are integers, then so are \(x^2-y^2\), \(2xy\), and \(x^2+y^2\).

However, \(4k^2+4k+257\) is not a perfect square trinomial, which means for any integer \(k\), the length of the hypotenuse is not an integer, so such a triangle doesn't exist.

MrRoyal

Pythagorean triples are used to determine if a triangle is a right triangle or not.

The legs of 16 units and an odd number cannot be a right triangle

The Pythagorean triples are given by:

\(x^2 - y^2\\\), \(2xy\) and \(x^2 + y^2\)

  • Given that one of the side lengths is 16 units.
  • The other side length can be represented as k + 1, where k is a non-negative even integer.

So, we have:

\(x^2 + y^2 = 16^2 + (k + 1)^2\)

Take square roots of both sides

\(\sqrt{x^2 + y^2} = \sqrt{16^2 + (k + 1)^2}\)

\(\sqrt{x^2 +y^2} = \sqrt{256 + k^2 + 2k + 1}\)

Collect like terms

\(\sqrt{x^2 + y^2} = \sqrt{256 + 1+ k^2 + 2k }\)

\(\sqrt{x^2 +y^2} = \sqrt{257+ k^2 + 2k }\)

Rewrite the above equation as:

\(\sqrt{x^2 +y^2} = \sqrt{k^2 + 2k + 257}\)

In the above equation, \(k^2 + 2k + 257\) cannot be expressed as a perfect square.

Hence, the lengths of 16 and an odd number cannot be a right triangle

Read more about Pythagorean triples at:

https://brainly.com/question/16314667

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