Formulate the Relationship Between Length and Width

Oct 13, 2025 by magicalceylontours.com 18 views

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Problem Statement

The area of a rectangle is given as $65 \, \text{ft}^2$ , and the length of the rectangle is 3 ft less than twice the width. We need to find the dimensions of the rectangle, specifically the length and width.

Let's Start with the Given Information

The area of the rectangle is $65 \, \text{ft}^2$ .
The length of the rectangle is 3 ft less than twice the width.

Define the Variables

Let's define the variables to represent the dimensions of the rectangle:

Width: $w$ ft
Length: $l$ ft

Formulate the Relationship Between Length and Width

According to the problem statement, the length of the rectangle is 3 ft less than twice the width. We can express this relationship as:

$l = 2w - 3$

Formulate the Equation for the Area of the Rectangle

The area of a rectangle is given by the formula:

$A = l \times w$

We are given that the area of the rectangle is $65 \, \text{ft}^2$ . Substituting the expression for $l$ from the previous equation, we get:

$65 = (2w - 3) \times w$

Expand and Simplify the Equation

Expanding the equation, we get:

$65 = 2w^2 - 3w$

Rearranging the terms, we get:

$2w^2 - 3w - 65 = 0$

Solve the Quadratic Equation

We can solve the quadratic equation using the quadratic formula:

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

In this case, $a = 2$ , $b = -3$ , and $c = -65$ . Plugging in these values, we get:

$w = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-65)}}{2(2)}$

Simplifying the expression, we get:

$w = \frac{3 \pm \sqrt{9 + 520}}{4}$

$w = \frac{3 \pm \sqrt{529}}{4}$

$w = \frac{3 \pm 23}{4}$

Find the Possible Values of Width

We have two possible values for the width:

$w = \frac{3 + 23}{4} = \frac{26}{4} = 6.5$

$w = \frac{3 - 23}{4} = \frac{-20}{4} = -5$

Since the width cannot be negative, we discard the negative value.

Find the Corresponding Length

Now that we have the width, we can find the corresponding length using the equation:

$l = 2w - 3$

Substituting $w = 6.5$ , we get:

$l = 2(6.5) - 3$

$l = 13 - 3$

$l = 10$

Conclusion

The dimensions of the rectangle are:

Width: $6.5$ ft
Length: $10$ ft

The area of the rectangle is indeed $65 \, \text{ft}^2$ , as given in the problem statement.

Discussion

This problem involves finding the dimensions of a rectangle given its area and a relationship between the length and width. We used algebraic techniques to solve the quadratic equation and find the possible values of width. By discarding the negative value, we obtained the correct width and corresponding length.

References

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Q: What is the formula for the area of a rectangle?

A: The formula for the area of a rectangle is:

$A = l \times w$

Where $A$ is the area, $l$ is the length, and $w$ is the width.

Q: How do I find the dimensions of a rectangle given its area and a relationship between the length and width?

A: To find the dimensions of a rectangle given its area and a relationship between the length and width, you can use the following steps:

Define the variables to represent the dimensions of the rectangle.
Formulate the relationship between the length and width.
Formulate the equation for the area of the rectangle.
Solve the quadratic equation to find the possible values of width.
Find the corresponding length using the equation.

Q: What is the relationship between the length and width of the rectangle in the problem statement?

A: The length of the rectangle is 3 ft less than twice the width. This can be expressed as:

$l = 2w - 3$

Q: How do I solve the quadratic equation to find the possible values of width?

A: To solve the quadratic equation, you can use the quadratic formula:

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

In this case, $a = 2$ , $b = -3$ , and $c = -65$ . Plugging in these values, you get:

$w = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-65)}}{2(2)}$

Simplifying the expression, you get:

$w = \frac{3 \pm \sqrt{9 + 520}}{4}$

$w = \frac{3 \pm \sqrt{529}}{4}$

$w = \frac{3 \pm 23}{4}$

Q: What are the possible values of width?

A: The possible values of width are:

$w = \frac{3 + 23}{4} = \frac{26}{4} = 6.5$

$w = \frac{3 - 23}{4} = \frac{-20}{4} = -5$

Since the width cannot be negative, we discard the negative value.

Q: What is the corresponding length?

A: Now that we have the width, we can find the corresponding length using the equation:

$l = 2w - 3$

Substituting $w = 6.5$ , we get:

$l = 2(6.5) - 3$

$l = 13 - 3$

$l = 10$

Q: What are the dimensions of the rectangle?

A: The dimensions of the rectangle are:

Width: $6.5$ ft
Length: $10$ ft

Q: How do I find the area of the rectangle?

A: To find the area of the rectangle, you can use the formula:

$A = l \times w$

Substituting the values of length and width, you get:

$A = 10 \times 6.5$

$A = 65$

Q: What is the final answer?

A: The final answer is:

Width: $6.5$ ft
Length: $10$ ft

The area of the rectangle is indeed $65 \, \text{ft}^2$ , as given in the problem statement.

Q: What are some related problems?

A: Some related problems include:

Finding the dimensions of a rectangle given its perimeter and area.
Solving quadratic equations with complex coefficients.
Using algebraic techniques to solve problems involving geometry and trigonometry.

Q: Where can I find more information?

A: You can find more information on the following resources: