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Jill has a rectangular poster hanging in her room. The poster’s width is 2.5 in. shorter than its length. The perimeter of the poster is 103 in.
Let L represent the poster’s length.

Answer 1

The length and area of a poster is required where the perimeter and width is given.

The length of the poster is 49 inches and the area is $122.5\ \text{in}^2$.

W = The width of the rectangle = 2.5 in.

P = The perimeter of the rectangle = 103 in.

L = Length of poster

Perimeter of a rectangle is given by

$P=2(L+W)\\\Rightarrow L=\dfrac{P}{2}-W\\\Rightarrow L=\dfrac{103}{2}-2.5\\\Rightarrow L=49\ \text{inches}$

The length of the poster is $49\ \text{inches}$.

The area of the poster is

$A=LW\\\Rightarrow A=49\times 2.5\\\Rightarrow A=122.5\ \text{in}^2$

The area of the poster is $122.5\ \text{in}^2$

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