Problem 29 - Algebra for College Students

Question

Solve each problem. The John Hancock Center in Chicago has a rectangular base. The length of the base measures \(65 \mathrm{ft}\) less than twice the width. The perimeter of the base is \(860 \mathrm{ft}\). What are the dimensions of the base?

Step-by-Step Solution

Answer

The dimensions are 165 feet by 265 feet.

Step 1: Define the Variables

Let the width of the base be denoted as \(w\) (in feet).

Step 2: Express the Length in Terms of the Width

The length of the base is 65 feet less than twice the width. Thus, the length can be expressed as \(2w - 65\) (in feet).

Step 3: Use the Perimeter Formula

The formula for the perimeter \(P\) of a rectangle is given by \(P = 2(l + w)\). Here, the perimeter is 860 feet. Substitute the length and width expressions into the formula: \[860 = 2((2w - 65) + w)\]

Step 4: Simplify and Solve for the Width

Expand and simplify the equation: \[860 = 2(2w - 65 + w)\] \[860 = 2(3w - 65)\] \[860 = 6w - 130\] Now, solve for \(w\): \[860 + 130 = 6w\] \[990 = 6w\] \[w = \frac{990}{6}\] \[w = 165\]

Step 5: Find the Length

Substitute the width back into the expression for the length: \[l = 2w - 65\] \[l = 2(165) - 65\] \[l = 330 - 65\] \[l = 265\]

Step 6: State the Dimensions

The dimensions of the base are 165 feet (width) and 265 feet (length).

Key Concepts

Solve Linear Equations

Solving linear equations is a key skill in algebra. It involves finding the value of a variable that makes the equation true. In the context of our problem, we had an equation for the perimeter of a rectangle that we needed to solve to find the dimensions.

Here's a quick breakdown:

• We start with the perimeter formula: \(P = 2(l + w)\).
• Plug in the expressions for the length \(l\) and width \(w\): \(860 = 2((2w - 65) + w)\).
• Expand and simplify the equation to isolate the variable: \(860 = 6w - 130\).
• Combine like terms and solve for \(w\): \(w = 165\).

By isolating the variable and solving the equation step by step, we find that the width \(w\) of the rectangular base is 165 feet. This process can be applied to any linear equation scenario.

Rectangular Dimensions

Understanding how to determine the dimensions of a rectangle from given information is critical in geometry. In the problem, we were given a perimeter and a relationship between length and width. Here’s how we approached it:

• We identified the relationship: the length \(l\) is 65 feet less than twice the width \(w\). This translates to the expression \((2w - 65)\) for the length.
• Next, we used the perimeter formula for rectangles: \P = 2(l + w)\.
• We substituted the relationship and perimeter value into the formula and solved for \w\.
• Finally, we used the width to calculate the length: \l = 2(165) - 65 = 265\.

The dimensions of the rectangular base thus calculated were 165 feet for the width and 265 feet for the length.

Variable Expressions

Variable expressions are algebraic expressions that include variables (letters that stand for unknown values), numbers, and arithmetic operations. In our exercise, we used variable expressions to represent the length and width of the rectangle:

• We set the width as \(w\) since it was a fundamental part of the given information.
• The length was described relative to the width, leading us to the expression \((2w - 65)\).
• These expressions allowed us to set up an equation based on the perimeter formula.

By defining variables and creating expressions based on given relationships, we simplify complex problems into solvable equations. This technique is very useful in algebra and beyond. With practice, setting up and solving variable expressions will become second nature.