Problem 116 - College Algebra with Modeling and Visualization

Question

Geometry A 174 -foot-long fence is being placed around the perimeter of a rectangular swimming pool that has a 3 -foot-wide sidewalk around it. The actual swimming pool without the sidewalk is twice as long as it is wide. Find the dimensions of the pool without the sidewalk.

Step-by-Step Solution

Answer

The pool's dimensions are 25 feet by 50 feet.

Step 1: Understand the Problem

We need to find the dimensions of the swimming pool without the sidewalk. We know that a fence is placed around the pool and sidewalk, which is 174 feet long, and the pool itself is twice as long as it is wide.

Step 2: Define Variables

Let's define the width of the swimming pool without the sidewalk as \( w \). Since the length is twice the width, the length of the pool will be \( 2w \). The sidewalk adds 3 feet around each side of the pool, increasing the total width and length by 6 feet (3 feet on each side for a total addition of 6 feet). Thus, the total width with the sidewalk is \( w + 6 \) and the total length is \( 2w + 6 \).

Step 3: Set Up the Equation for the Perimeter

The perimeter of the rectangular area covered by the fence is given by the sum of all its sides. Therefore, we have the equation: \[ 2(w + 6) + 2(2w + 6) = 174 \] This equation represents the total perimeter of the sidewalk and pool surrounded by the fence.

Step 4: Solve the Equation

Simplify the equation to find \( w \):\[ 2w + 12 + 4w + 12 = 174 \]\[ 6w + 24 = 174 \]Subtract 24 from both sides:\[ 6w = 150 \]Divide by 6:\[ w = 25 \]

Step 5: Find the Dimensions

Using \( w = 25 \), calculate the dimensions of the pool:- Width = \( w = 25 \) feet - Length = \( 2w = 50 \) feet

Step 6: Verify the Solution

Calculate the total perimeter using the dimensions including the sidewalk to ensure it adds up correctly:Width with sidewalk = \( 25 + 6 = 31 \) feetLength with sidewalk = \( 50 + 6 = 56 \) feetPerimeter: \[ 2(31) + 2(56) = 62 + 112 = 174 \] Since it matches the given perimeter, our dimensions are verified.

Key Concepts

Understanding Geometry Basics

Geometry involves studying the size, shape, and relative position of figures, along with the properties of space. When working with rectangles, there are a few key aspects to understand:

• A rectangle has two pairs of opposite sides that are equal in length.
• Angles within a rectangle are right angles, meaning they are each 90 degrees.
• The length and width define the shape and size of the rectangle, leading to the calculation of other properties like area and perimeter.

For our problem, understanding these basic geometric properties helps in forming equations that relate to the dimensions and the perimeter of rectangular shapes.

Calculating the Perimeter

The perimeter of a rectangle is the total distance around the edge of the rectangle. To find it, add up the lengths of all four sides. The formula for the perimeter \( P \) of a rectangle given length \( l \) and width \( w \) is:

• \( P = 2l + 2w \)

In our example, the perimeter includes the swimming pool and the surrounding sidewalk. Calculating the perimeter helps us understand the full length of the fencing needed. For a rectangle, make sure to add the additional lengths added by the sidewalks, which is crucial in problems involving a surrounding border.

Understanding Rectangular Dimensions

Rectangular dimensions refer to the length and width of a rectangle. In algebraic terms, if one dimension depends on the other, you can express one variable in terms of the other. In our problem:

• The width of the pool is represented as \( w \).
• The length is twice the width, represented as \( 2w \).
• Including the sidewalk, the total width becomes \( w + 6 \), and the total length becomes \( 2w + 6 \).

These expressions allow us to set up equations that properly describe all aspects of the problem, including how the dimensions change when the sidewalk is added.

Solving with Algebraic Equations

Algebraic equations allow us to find unknown values by setting up equalities that represent real-world situations. In our problem, we solve for \( w \), the width of the pool, by:

• Setting up the equation based on the perimeter:\[ 2(w + 6) + 2(2w + 6) = 174 \]
• Simplifying the equation:\[ 6w + 24 = 174 \]
• Solving the equation to find \( w \): subtract 24, then divide by 6 to arrive at \( w = 25 \).

Once you have the width, finding the length becomes straightforward, as it is twice the width. Using algebraic equations provides a systematic way to find dimensions in complex geometric problems.