If you have 36 ft of fencing, what are the bases and side lengths of the different parallelograms you could enclose with the fencing? Consider only whole-number dimensions.

Respuesta :

Answer:

[tex](l,w) = (1,17)[/tex]

[tex](l,w) = (2,16)[/tex]

[tex](l,w) = (3,15)[/tex]

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[tex](l,w) = (9,9)[/tex]

Step-by-step explanation:

Given

[tex]P = 36ft[/tex] --- perimeter

[tex]l \to length[/tex]

[tex]w \to width[/tex]

Required

Possible dimension of different parallelogram

The perimeter is calculated as:

[tex]P=2(l+w)\\\\\\[/tex]

So,we have:

[tex]2(l+w)=36[/tex]

Divide by 2

[tex]l + w = 18[/tex]

Since l and w must be positive integers, the possible dimensions are:

[tex](l,w) = (1,17)[/tex]

[tex](l,w) = (2,16)[/tex]

[tex](l,w) = (3,15)[/tex]

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[tex](l,w) = (9,9)[/tex]