4.8 Polynomial Word Problems. a) Define the variable, b) Write the ... - Free Printable

Problem Statement:

The length of a rectangle is 11 cm more than its width. The area of the rectangle is 2040 cm². Find the dimensions of the rectangle.

Solution:

#### Step 1: Define Variables
Let:
- \( W \) = width of the rectangle (in cm)
- \( L \) = length of the rectangle (in cm)

From the problem, we know:
1. The length is 11 cm more than the width:
\[
L = W + 11
\]
2. The area of the rectangle is 2040 cm²:
\[
\text{Area} = L \cdot W = 2040
\]

#### Step 2: Substitute the Expression for \( L \) into the Area Formula
Using \( L = W + 11 \), substitute this into the area formula:
\[
L \cdot W = 2040
\]
\[
(W + 11) \cdot W = 2040
\]

#### Step 3: Expand and Rearrange the Equation
Expand the left-hand side:
\[
W^2 + 11W = 2040
\]
Rearrange the equation to form a standard quadratic equation:
\[
W^2 + 11W - 2040 = 0
\]

#### Step 4: Solve the Quadratic Equation
We solve the quadratic equation \( W^2 + 11W - 2040 = 0 \) using factoring. We need to find two numbers that multiply to \(-2040\) and add up to \(11\).

Factorize the quadratic equation:
\[
W^2 + 11W - 2040 = (W + 51)(W - 40) = 0
\]

Set each factor equal to zero:
\[
W + 51 = 0 \quad \text{or} \quad W - 40 = 0
\]
\[
W = -51 \quad \text{or} \quad W = 40
\]

Since the width \( W \) cannot be negative, we discard \( W = -51 \). Thus:
\[
W = 40
\]

#### Step 5: Find the Length
Using the relationship \( L = W + 11 \):
\[
L = 40 + 11 = 51
\]

#### Step 6: Verify the Solution
Check the area:
\[
L \cdot W = 51 \cdot 40 = 2040 \quad \text{(which matches the given area)}
\]

#### Final Answer:
The dimensions of the rectangle are:
\[
\boxed{40 \text{ cm} \times 51 \text{ cm}}
\]