Problem 269 - Elementary Algebra

Question

In the following exercises, solve using rectangle properties. The perimeter of a rectangle is 150 feet. The length of the rectangle is twice the width. Find the length and width of the rectangle.

Step-by-Step Solution

Answer

The width is 25 feet and the length is 50 feet.

Step 1: Write down the formulas

The perimeter of a rectangle is given by the formula N N P = 2L + 2W, where L is the length and W is the width. Given: Perimeter (P) = 150 feet and Length (L) is twice the Width (W), so L = 2W.

Step 2: Substitute given values into the formula

Replace L with 2W in the perimeter formula: N N N N150 = 2(2W) + 2W

Step 3: Simplify and solve for W

Combine like terms and solve for W: N N N N N N 150 = 4W + 2W 150 = 6W Divide both sides by 6 to find W W = 25

Step 4: Find the length

Use W to find L using L = 2W: N N L= 2(25) = 50

Step 5: Verify the result

Check the values by substituting them back into the perimeter formula: N P = 2L + 2W = 2(50) + 2(25) = 100 + 50 = 150 feet, which matches the given perimeter.

Key Concepts

rectangle properties

Rectangles are simple geometric shapes often encountered in algebra problems. They have some important properties: all angles are right angles (90 degrees), and opposite sides are parallel and equal in length. Understanding these properties can help you solve problems related to the perimeter and area of rectangles. The perimeter of a rectangle is the total distance around the edge of the rectangle. Other key attributes are the length (the longer side) and the width (the shorter side). To solve a problem involving rectangles, you need to know these properties.

perimeter formula

The perimeter of a rectangle can be found using a simple formula. This formula helps you calculate the total distance around the rectangle. The formula is:
\(P = 2L + 2W\), where \(P\) is the perimeter, \(L\) is the length, and \(W\) is the width. This formula combines all the sides of the rectangle. Two lengths and two widths add up to the total perimeter.
In the exercise, it is given that the perimeter is 150 feet, and the length is twice the width. By substituting these values into the formula, we can find out the actual dimensions of the rectangle.

algebraic equations

Solving rectangle problems often involves algebraic equations. Let's break down how we used algebra here:

• First, we wrote the perimeter formula: \(P = 2L + 2W\).
• Then, we used the given conditions: \(P = 150\) and \(L = 2W\).
• We substituted \(L = 2W\) into the perimeter formula: \(150 = 2(2W) + 2W\).
• We simplified the equation: \(150 = 4W + 2W\).
• This gave us \(6W = 150\).
• Dividing both sides by 6, we solved for the width: \(W = 25\).

Using \(W = 25\), we found the length: \(L = 2(25) = 50\). This shows how algebra can be used to solve real-world problems by setting up equations based on given information.
The final verification step confirms our solution by substituting the values of \(L\) and \(W\) back into the original formula to ensure the calculated perimeter matches the given perimeter.