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A rectangle school banner has a length of 44 inches, a perimeter of 156 inches , and an area of 1496 square inches. The cheerleaders make signs similar to the banner. The length of a sign is 11 inches. What is its perimeter and its area ?

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DementedConvict
Since you have the length and the area, you can assume the width

1496/44 = Width              Because  Area = L * W

Width = 34

So this is a ratio problem

44/11 = ratio

So, the ratio is 4:1
So since we have the width, we can divide the width of the school banner by 4 because of the scale

34/4 = 8.5 = width of the sign

The area of the sign is 8.5*11 = 93.5

and the perimeter is just that,

8.5 * 2 + 11*2 = 39

Perimeter = 39 inches

Area = 93.5 inches²
carlosego

For the school banner we have:

[tex] A = w * l
[/tex]

Where,

w: width of the banner

l: length of the banner

Substituting values we have:

[tex] 1496 = w * 44
[/tex]

Clearing the width we have:

[tex] w = \frac{1496}{44}

w = 34
[/tex]

Then, for the cheerleaders we have:

The ratio of measures is:

[tex] \frac{l}{l'} = \frac{44}{11} = 4
[/tex]

Therefore, the width is given by:

[tex] w '= \frac{w}{4}
[/tex]

[tex] w '= \frac{34}{4}

w '= 8.5
[/tex]

Then, the perimeter is given by:

[tex] P = 2l '+ 2w'
[/tex]

Substituting values we have

[tex] P = 2 (11) + 2 (8.5)

P = 22 + 17

P = 39
[/tex]

The area is given by:

[tex] A = w '* l'
[/tex]

Substituting values we have

[tex] A = (8.5) * (11)

A = 93.5
[/tex]

Answer:

Its perimeter and its area are:

P = 39 inches

A = 93.5 square inches