Teorema de pitágoras. Las medidas de los lados y la diagonal de un rectángulo son tres números impares consecutivos. Hallar los valor

Question

Teorema de pitágoras. Las medidas de los lados y la diagonal de un rectángulo son tres números impares consecutivos.
Hallar los valores de estos elementos.

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Valentina 3 months 2021-10-19T20:10:35+00:00 1 Answer 0 views 0

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    2021-10-19T20:11:51+00:00

    Answer:

    This question says:

    “the lenghts of the sides and the diagonal of a rectangle are 3 odd consecutive numbers”.

    This means that if x is an odd number, then the meassures are:

    x, x + 2, x + 4.

    We add 2 in each case, because the next consecutive number of, for example, 3 is 3 + 2 = 5, and so on.

    Now, from the Pitagoras theorem we know that, for a triangle rectangle, we have

    A^2 + B^2 = H^2

    where A and B are the cathetus, and H is the hypotenuse.

    In our case the smaller odd numbers will be the cathetus and the larger one will be the hypotenuse.

    x^2 + (x + 2)^2 = (x + 4)^2

    x^2 + x^2 + 4x + 4 = x^2 + 8x + 16

    x^2 – 4x -12 = 0

    now we need to solve this for x

    we know that for a quadratic equation of the form:

    ax^2 + bx + c = 0

    the solutions are:

    x = \frac{-b +- \sqrt{b^2 -4ac} }{2a}

    in our case we have:

    x = \frac{4 +- \sqrt{16 +48} }{2} = \frac{4+-8}{2}

    The only positive value is x = 12/2 = 6

    but 6 is not odd, so we can never have that the sides and the diagonal are odd consecutive numbers.

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45:7+7-4:2-5:5*4+35:2 =? ( )