Question

1. the length of a rectangle is 1 foot more than twice the width. the area is 55 square feet. find the dimensions of the rectangle. let l = a = l w show set up, then solve. ( )( ) = 0

Explanation:

Step1: Define variables

Let \( W \) be the width of the rectangle. Then the length \( L = 2W + 1 \).

Step2: Use the area formula

The area formula \( A = LW \), and \( A = 55 \). Substitute \( L \) into the formula: \( 55=(2W + 1)W \).

Step3: Expand the equation

Expand \( (2W + 1)W \) to get \( 2W^{2}+W - 55=0 \).

Step4: Factor the quadratic equation

Factor \( 2W^{2}+W - 55 \). We need two numbers \( a \) and \( b \) such that \( a\times b=2\times(- 55)=-110 \) and \( a + b = 1 \). The numbers are \( 11 \) and \( - 10 \).
Rewrite the middle term: \( 2W^{2}+11W-10W - 55 = 0 \).
Group terms: \( W(2W + 11)-5(2W + 11)=0 \).
Factor out \( (2W + 11) \): \( (2W + 11)(W - 5)=0 \).

Step5: Solve for \( W \)

Set each factor equal to zero:

• \( 2W+11 = 0\Rightarrow W=-\frac{11}{2} \) (rejected since width cannot be negative).
• \( W - 5=0\Rightarrow W = 5 \).

Step6: Find the length

If \( W = 5 \), then \( L=2W + 1=2\times5+1 = 11 \).

Answer:

The width of the rectangle is \( 5 \) feet and the length is \( 11 \) feet.