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The width of the rectangle is 5 inches shorter than the length. The perimeter of the rectangle can be written as a function of the length p(l)=______. The area of the rectangle can be written as a function of the length a(l)= _____

Sagot :

Answer:

P = 4L - 10

A = [tex]\rm L^2-5L[/tex]

Step-by-step explanation:

Area & Perimeter

Area

By taking the product of a 2-D shape's side lengths, the area calculates how much space a 2-D shape takes up.

Perimeter

A 2-D shape's perimeter is the sum of all its side lengths.

[tex]\hrulefill[/tex]

Solving the Problem

Width

We're told that the width is 5 less than the length.

If the measurement of the width is W, and the measurement of the length is L inches then,

                                          W = L - 5.

Perimeter

The side lengths of a rectangle are 2 with a measurement of the width and 2 with length.

So,

                              P = L + L + W + W or 2(L + W)

                              P = 2[L + (L - 5)} = 2(2L - 5)

                                         P = 4L - 10

Area

The area of a rectangle is the product of its width and length.

So,

                              A = LW = L(L - 5) = [tex]\rm L^2-5L[/tex].

The perimeter of the rectangle can be written as a function of the length p(l)= 4l - 10. The area of the rectangle can be written as a function of the length a(l)= [tex]\underline{l^2 - 5l}[/tex].

Let l represent the length of the rectangle.

Since the width is 5 inches shorter than the length, let w = l - 5.

Perimeter Function

  • The perimeter P of a rectangle is given by the formula:
    [tex]P = 2l + 2w[/tex]

  • Substituting w:
    [tex]P(l) = 2l + 2(l - 5)[/tex]

  • Simplifying this:
    [tex]P(l) = 2l + 2l - 10 = 4l - 10[/tex]

Area Function:

  • The area A of a rectangle is given by the formula:
    [tex]A=l\cdot w[/tex]

  • Substituting w:
    [tex]A(l) = l \cdot (l - 5)[/tex]

  • Simplifying this:
    [tex]A(l) = l^2 - 5l[/tex].
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