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Hosn is a picture framer. He is framing a picture that is 24 inches by 32 inches with a
frame that is 2.5 inches wide. What is the outer perimeter of the framed picture In yards, feet, and inches.​

Respuesta :

gilbertazavedo72

Answer: 122 sq. in.

Step-by-step explanation:

P = 2 (l + w)

P = 2 (34.5 + 26.5)

P = 2 x 61

P = 122 sq. in. = 0.847 sq ft = 0.0941 sq. yards

MrRoyal

Perimeter is the sum of the side lengths of a shape. The outer perimeter of the framed picture is:

  • 132 inches
  • 3.67 yard
  • 11 feet

Given that:

[tex]Length = 24in\\Width = 32in[/tex]

The width of the frame is given as:

[tex]Frame =2.5in[/tex]

So, the actual length and width of the framed picture is:

[tex]Length =24 + 2.5 + 2.5[/tex]

[tex]Length =29in[/tex]

[tex]Width = 32 + 2.5 + 2.5[/tex]

[tex]Width = 37in[/tex]

The outer perimeter (P) in inches is:

[tex]P= 2 \times (Length + Width)[/tex]

So, we have:

[tex]P= 2 \times (29in + 37in)[/tex]

[tex]P= 132in[/tex]

The perimeter in yard, is as follows:

[tex]P= \frac{132}{36}yd[/tex]

[tex]P = 3.67yd[/tex]

The perimeter in feet is as follows:

[tex]P= \frac{132}{12}ft[/tex]

[tex]P = 11ft[/tex]

Hence, the outer perimeter of the frame is 132 inches

Read more about perimeters at:

https://brainly.com/question/6465134

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