The length of a rectangle is 3 meters less than twice the width. If the area of the rectangle is 209 square​ meters, find the dimensions.

To solve this problem, let's denote the width of the rectangle as \( w \) meters. According to the problem, the length of the rectangle, \( l \), is 3 meters less than twice the width. This can be written as:

\[
l = 2w - 3
\]

The area of a rectangle is given by the formula:

\[
\text{Area} = l \times w
\]

We are given that the area of the rectangle is 209 square meters. Substituting the expressions for the length and area, we get:

\[
209 = (2w - 3) \times w
\]

Now, let's solve this quadratic equation for \( w \).

### Step-by-Step Solution

1. **Expand and simplify the equation:**

\[
209 = (2w - 3)w
\]

\[
209 = 2w^2 - 3w
\]

2. **Rearrange the equation to standard quadratic form:**

\[
2w^2 - 3w - 209 = 0
\]

3. **Solve the quadratic equation using the quadratic formula:**

The quadratic formula is:

\[
w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For our equation, \(a = 2\), \(b = -3\), and \(c = -209\). Substituting these values into the quadratic formula gives:

\[
w = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-209)}}{2 \cdot 2}
\]

\[
w = \frac{3 \pm \sqrt{9 + 1672}}{4}
\]

\[
w = \frac{3 \pm \sqrt{1681}}{4}
\]

4. **Calculate the square root and find the two possible solutions for \( w \):**

\[
w = \frac{3 \pm 41}{4}
\]

\[
w_1 = \frac{3 + 41}{4} = \frac{44}{4} = 11
\]

\[
w_2 = \frac{3 - 41}{4} = \frac{-38}{4} = -9.5
\]

Since the width \( w \) cannot be negative, we discard \( w_2 = -9.5 \).

Thus, \( w = 11 \) meters.

5. **Find the length \( l \):**

\[
l = 2w - 3 = 2(11) - 3 = 22 - 3 = 19
\]

### Conclusion

The dimensions of the rectangle are:
- **Width (w):** 11 meters
- **Length (l):** 19 meters

Would you like more details or have any questions?

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**Related Questions:**

1. What is the perimeter of this rectangle?
2. If the width was increased by 5 meters, what would be the new area of the rectangle?
3. How would the dimensions change if the area were doubled?
4. What are the dimensions of a square with the same area as the rectangle?
5. What is the length-to-width ratio of this rectangle?

**Tip:** Always check if the dimensions make sense in the context of the problem (e.g., dimensions should not be negative).

Learn how to find the dimensions of a rectangle when given its area of 209 square meters. This problem involves algebraic manipulation and quadratic equations, specifically the area formula for a rectangle. Suitable for Grades 7-9 students.

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