Question

The perimeter of a rectangular floor is 90 feet. Find the dimensions of the floor if the length is twice the width.

Short answer

The width is 15 feet and the length is 30 feet.

Step-by-step solution

Understand the Problem

Given a rectangular floor with a perimeter of 90 feet, and the length is twice the width. We need to find the dimensions (length and width) of the floor.

Define Variables

Let the width of the rectangle be denoted by \(w\). Since the length is twice the width, the length can be denoted by \(2w\).

Recall the Perimeter Formula

The perimeter \(P\) of a rectangle is given by the formula: \[ P = 2(l + w) \] where \(l\) is the length and \(w\) is the width.

Substitute Known Values

Substitute \(P = 90\), \(l = 2w\), and \(w\) into the perimeter formula: \[ 90 = 2(2w + w) \]

Simplify the Equation

Simplify the equation: \[ 90 = 2(3w) \] \[ 90 = 6w \]

Solve for the Width

Divide both sides by 6 to find the width: \[ w = \frac{90}{6} \] \[ w = 15 \text{ feet} \]

Find the Length

Since the length is twice the width, calculate the length: \[ l = 2w = 2 \times 15 = 30 \text{ feet} \]

Verify the Solution

Verify by checking the perimeter: \[ P = 2(l + w) = 2(30 + 15) = 2(45) = 90 \text{ feet} \]. The solution is correct.

Key concepts

rectangular perimeter formula

The rectangular perimeter formula is essential to solve many geometry problems. The perimeter is the total distance around the outside of the rectangle. To calculate it, we use the formula: \[ P = 2(l + w) \]

where \( P \) is the perimeter, \( l \) is the length, and \( w \) is the width.

This formula works because a rectangle has two pairs of equal sides. By adding the length and width together and then doubling the result, we account for all four sides.

Understanding this formula helps in analyzing and solving rectangles' dimensions in various scenarios, aiding particularly in area and perimeter calculations seamlessly.

To solidify this concept, you can practice by finding the perimeters of rectangles with different lengths and widths, substituting them into the formula and verifying the results.

variable substitution

Variable substitution is a fundamental concept in algebra that helps simplify and solve equations. In this exercise, we use it to express unknown dimensions in terms of a known quantity.

For instance, let's denote the width of the rectangle as \( w \). Since the length is twice the width, we can represent the length as \( 2w \). This substitution is powerful as it transforms the perimeter formula:

\[ P = 2(l + w) \rightarrow P = 2(2w + w) \]

Such substitution converts the initial problem into a single-variable equation, making it easier to handle and solve.

Practicing variable substitution in various mathematical scenarios can help in understanding how to break down complex problems into simpler, manageable parts. This technique is widely used in geometry, algebra, and calculus.

solving linear equations

Solving linear equations is a critical skill in mathematics that allows us to find unknown values. In this case, the equation is derived from the perimeter formula, with all substitutions made:

\[ 90 = 2(3w) \]

Simplifying the equation, we get:

\[ 90 = 6w \]

Next, we solve for the width \( w \) by isolating it on one side of the equation:

\[ w = \frac{90}{6} = 15 \text{ feet} \]

After figuring out the width, finding the length is straightforward:

\[ l = 2w = 2 \times 15 = 30 \text{ feet} \]

This step-by-step process emphasizes the importance of simplifying and isolating variables in linear equations. Regular practice will solidify these techniques and make solving such equations second nature.

Try solving different linear equations by isolating variables and using similar methods, which will help reinforce the concept.