The Area Of A Rectangle Is $35 \, \text{ft}^2$, And The Length Of The Rectangle Is 8 Ft Less Than Three Times The Width. Find The Dimensions Of The Rectangle.Length: $\square$ FtWidth: $\square$ Ft

Introduction

In this article, we will explore the problem of finding the dimensions of a rectangle given its area and a relationship between its length and width. The area of a rectangle is given by the formula $A = lw$, where $A$ is the area, $l$ is the length, and $w$ is the width. We will use this formula to find the dimensions of the rectangle.

Problem Statement

The area of a rectangle is $35 \, \text{ft}^2$, and the length of the rectangle is 8 ft less than three times the width. We need to find the dimensions of the rectangle.

Let's Start with the Given Information

• The area of the rectangle is $35 \, \text{ft}^2$.
• The length of the rectangle is 8 ft less than three times the width.

Mathematical Representation

Let's represent the width of the rectangle as $w$. Then, the length of the rectangle is $3w - 8$. We can use the formula for the area of a rectangle to set up an equation:

$$lw = 35$$

Substituting the expressions for length and width, we get:

$$(3w - 8)w = 35$$

Expanding and Simplifying the Equation

Expanding the left-hand side of the equation, we get:

$$3w^2 - 8w = 35$$

Rearranging the terms, we get:

$$3w^2 - 8w - 35 = 0$$

Solving the Quadratic Equation

We can solve this quadratic equation using the quadratic formula:

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

In this case, $a = 3$, $b = -8$, and $c = -35$. Plugging these values into the formula, we get:

$$w = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(3)(-35)}}{2(3)}$$

Simplifying the expression, we get:

$$w = \frac{8 \pm \sqrt{64 + 420}}{6}$$

$$w = \frac{8 \pm \sqrt{484}}{6}$$

$$w = \frac{8 \pm 22}{6}$$

Finding the Width

We have two possible values for the width:

$$w = \frac{8 + 22}{6} = \frac{30}{6} = 5$$

$$w = \frac{8 - 22}{6} = \frac{-14}{6} = -\frac{7}{3}$$

Since the width cannot be negative, we discard the second solution.

Finding the Length

Now that we have found the width, we can find the length:

$$l = 3w - 8 = 3(5) - 8 = 15 - 8 = 7$$

Conclusion

In this article, we have found the dimensions of a rectangle given its area and a relationship between its length and width. The width of the rectangle is 5 ft, and the length is 7 ft.

Final Answer

The dimensions of the rectangle are:

• Length: 7 ft
• Width: 5 ft
  The Area of a Rectangle: Q&A =============================

Introduction

In our previous article, we explored the problem of finding the dimensions of a rectangle given its area and a relationship between its length and width. We used the formula for the area of a rectangle to set up an equation and solved it to find the dimensions of the rectangle. In this article, we will answer some common questions related to the problem.

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

A: The formula for the area of a rectangle is $A = lw$, 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 its length and width?

A: To find the dimensions of a rectangle given its area and a relationship between its length and width, you can use the formula for the area of a rectangle to set up an equation. Let's say the length of the rectangle is 8 ft less than three times the width. We can represent the width as $w$ and the length as $3w - 8$. We can then use the formula for the area of a rectangle to set up an equation:

$$(3w - 8)w = 35$$

Solving this equation will give us the value of $w$, which is the width of the rectangle. We can then find the length by substituting the value of $w$ into the expression for the length.

Q: What if the relationship between the length and width is more complex?

A: If the relationship between the length and width is more complex, you may need to use a different approach to solve the problem. For example, if the length is 2 ft more than twice the width, you can represent the width as $w$ and the length as $2w + 2$. You can then use the formula for the area of a rectangle to set up an equation:

$$(2w + 2)w = 35$$

Solving this equation will give you the value of $w$, which is the width of the rectangle. You can then find the length by substituting the value of $w$ into the expression for the length.

Q: Can I use the quadratic formula to solve the equation?

A: Yes, you can use the quadratic formula to solve the equation. The quadratic formula is:

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

In this case, $a = 3$, $b = -8$, and $c = -35$. Plugging these values into the formula, we get:

$$w = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(3)(-35)}}{2(3)}$$

Simplifying the expression, we get:

$$w = \frac{8 \pm \sqrt{64 + 420}}{6}$$

$$w = \frac{8 \pm \sqrt{484}}{6}$$

$$w = \frac{8 \pm 22}{6}$$

Q: What if I get two solutions for the width?

A: If you get two solutions for the width, you need to check which one is valid. In this case, we got two solutions:

$$w = \frac{8 + 22}{6} = \frac{30}{6} = 5$$

$$w = \frac{8 - 22}{6} = \frac{-14}{6} = -\frac{7}{3}$$

Since the width cannot be negative, we discard the second solution.

Q: Can I use this method to find the dimensions of any rectangle?

A: Yes, you can use this method to find the dimensions of any rectangle given its area and a relationship between its length and width. However, you need to make sure that the relationship between the length and width is linear, meaning that it can be represented by a straight line.

Conclusion

In this article, we have answered some common questions related to finding the dimensions of a rectangle given its area and a relationship between its length and width. We have shown how to use the formula for the area of a rectangle to set up an equation and solve it to find the dimensions of the rectangle. We have also discussed how to use the quadratic formula to solve the equation and how to check which solution is valid.