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.

Answer 1

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 = (-b +- √(b^2 -4ac) )/(2a)`

in our case we have:

`x = (4 +- √(16 +48) )/(2) = (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.