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The length of a rectangular poster is 5 more inches than half its width. The area of the poster is 48 square inches. Solve for the dimensions (length and width) of the poster.

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Answer:

the dimensions of the rectangular poster are width = 6 inches and length = 8 inches.

Explanation:

Let's assume the width of the rectangular poster is represented by 'w' inches.

According to the given information, the length of the poster is 5 more inches than half its width. So, the length can be represented as (0.5w + 5) inches.

The formula for the area of a rectangle is given by:

Area = length * width

We are given that the area of the poster is 48 square inches, so we can set up the equation:

(0.5w + 5) * w = 48

Now, let's solve this equation to find the value of 'w' (width) first:

0.5w^2 + 5w = 48

Multiplying through by 2 to eliminate the fraction:

w^2 + 10w - 96 = 0

Now, we can factorize this quadratic equation:

(w - 6)(w + 16) = 0

Setting each factor to zero:

w - 6 = 0 or w + 16 = 0

Solving for 'w', we get:

w = 6 or w = -16

Since the width of a rectangle cannot be negative, we discard the value w = -16.

Therefore, the width of the poster is 6 inches.

To find the length, we substitute the value of the width (w = 6) into the expression for the length:

Length = 0.5w + 5 = 0.5 * 6 + 5 = 3 + 5 = 8 inches

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User FrankV
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