Problem 149 - College Algebra

Question

A pool measuring 10 meters by 20 meters is surrounded by a path of uniform width, as shown in the figure at the top of the next column. If the area of the pool and the path combined is 600 square meters, what is the width of the path?

Step-by-Step Solution

Answer

The width of the path is 5 meters.

Step 1: Determine the dimensions of the pool and path

The dimensions of the pool are 10 meters x 20 meters. The path is of uniform width, let's denote it with a variable \( w \). Therefore, the total length and width of the pool plus path would be (10+2w) and (20+2w).

Step 2: Formulate an equation for the total area

As we know, the area \( A \) of the rectangle is length x width. Therefore, the equation for the area of the pool plus path is (10+2w) * (20+2w) = 600.

Step 3: Solve the equation

This is a quadratic equation. By simplifying the equation we will obtain 4w^2 + 60w - 400 = 0. Solving this quadratic equation, we find two values, \( w = 5 \) and \( w = -20 \). Since the width cannot be negative, \( w = 5 \) meters.

Key Concepts

Area Calculation

Solving problems involving area calculation requires understanding how dimensions multiply to create a space within boundaries. Consider a pool that measures 10 meters by 20 meters.

• The pool's area is calculated by multiplying its length and width: \( 10 \times 20 = 200 \text{ square meters} \).
• When a path surrounds the pool, the total area includes both the pool and the path.
• The task is to find the combined area of the pool and path, given as 600 square meters.

The problem involves first understanding how these separate areas combine and then employing algebra to solve for unknown dimensions. This approach requires a solid grasp of both basic multiplication and how these numbers relate to spatial dimensions.

Uniform Width

In geometry problems, uniform width refers to a consistent measurement extending from the sides of an object. Here, the path around the pool has a uniform width \( w \).

• Consider the pool: 10 meters by 20 meters.
• Add the path: its width \( w \) adjusts both the length and width uniformly.

This adjustment means each side of the pool enlarges by \( 2w \) since the path is present on all sides. Consequently, the new dimensions of the entire area, including the path, become \( (10 + 2w) \) by \( (20 + 2w) \). Understanding this concept helps you correctly formulate how dimensions change in a problem.

Problem Solving

In solving problems like this one, strategic formulation of an equation is key. The task is to determine the width \( w \) using the total area equation.

• Known total area: 600 square meters.
• Represented by: \( (10 + 2w)(20 + 2w) = 600 \).

The equation models all the information given in the exercise. This transforms the problem into a solvable mathematical challenge using the algebraic manipulation of expressions. Once simplified, you achieve: \( 4w^2 + 60w - 400 = 0 \), a quadratic equation, where you'd apply the quadratic formula or factorization to find \( w \. \)

Dimension Analysis

Dimension analysis involves understanding how changes in one dimension affect the whole. In this problem, the width of the path alters the dimensions of the pool.
Here's the process:

• Original dimensions: 10 by 20 meters.
• Path added, extending each side by \( w \).
• New dimensions: \( 10 + 2w \) by \( 20 + 2w \).

The process of expanding dimensions essentially calculates how the path's width integrates into the original layout to increase the area. Once the quadratic equation is set, solving it reveals one viable solution for \( w \), ensuring consistency with the constraints and conditions provided.