Question

Geometry A rectangular pool is 30 feet wide and 40 feet long. It is surrounded on all four sides by a wooden deck that is \(x\) feet wide. The total area enclosed within the perimeter of the deck is 3000 square feet. What is the width of the deck?

Short answer

The width of the wooden deck is 10 feet.

Step-by-step solution

Identify given parameters

The width and length of the rectangular pool are given, being 30 feet and 40 feet respectively. The total area enclosed by the deck is given as 3000 square feet.

Set up the equation

The area of a rectangular space enclosed by the wooden deck is length times width. In this case, the length and width of the overall rectangle are both increased by \(2x\) because the deck surrounds all four sides of the pool. Therefore, the equation to represent the total area is \((30+2x) (40+2x) = 3000\).

Simplify and solve the equation

First, expand the brackets to get \(1200 + 140x + 4x^2 = 3000\). Rearrange the terms and subtract 3000 from both sides to get a quadratic equation set to zero: \(4x^2 + 140x - 1800 = 0\). Divide the whole equation by 4 to simplify it, giving \(x^2 + 35x - 450 = 0\). Factorize the equation to find \(x (x + 45) - 10 (x + 45) = 0\). This gives \((x - 10)(x + 45) = 0\). So, \(x = 10 \ or \ x = -45\). However, in the context of this problem, \(x\) cannot be negative. Therefore, the width of the wooden deck is 10 feet.

Key concepts

Quadratic Equations

Quadratic equations are a key concept in algebra and are commonly used in various mathematical problems, including geometry and physics. A quadratic equation is any equation that can be arranged in the form \( ax^2 + bx + c = 0 \), where \(a\), \(b\), and \(c\) are constants, and \(x\) represents the variable. Understanding how to manipulate and solve these equations is crucial.

When solving a quadratic equation, you often deal with:

• The standard form: simplifying the equation until you have all terms on one side and zero on the other, like in \(4x^2 + 140x - 1800 = 0\).
• Factoring: looking for two numbers that multiply to \( ac \) and add to \( b \), as we did with \((x - 10)(x + 45) = 0\).
• The quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), useful when factoring is challenging or impossible.

In the context of the exercise, quadratic equations help us determine potential measurements that solve spatial problems efficiently, avoiding guesswork.

Rectangular Area

Understanding the area of a rectangle is fundamental in geometry. The area is calculated by multiplying the rectangle's length by its width. For example, a pool that is 30 feet wide and 40 feet long has an area of \(30 \times 40 = 1200\) square feet.

The problem becomes more engaging when we introduce additional elements such as a surrounding deck. When a deck surrounds a rectangle, it affects the total area because it adds an increment (denoted as \(x\) feet in this exercise) to both the length and width.

• Overall width includes the pool's width plus twice the deck's width: \(30 + 2x\).
• Overall length includes the pool's length plus twice the deck's width: \(40 + 2x\).

Multiplying these new dimensions gives the total rectangular area enclosed: \((30 + 2x)(40 + 2x)\). Understanding how to modify basic area calculations with these additions is essential in geometry problem-solving.

Problem Solving

The skill set involved in problem-solving is critical, particularly when dealing with multi-layered mathematical problems. This exercise illustrates a typical scenario where algebraic and geometric principles merge to produce a solution.

Effective problem-solving involves:

• Identifying known parameters and variables, such as the dimensions of the pool and the total area including the deck.
• Setting up an appropriate equation by translating the word problem into a mathematical model. In this case, this leads to the quadratic equation \((30+2x)(40+2x) = 3000\).
• Simplifying and solving the equation through expansion, rearrangement, and solving for \(x\). Here simplification led to \(x^2 + 35x - 450 = 0\) which was factored accordingly.
• Ensuring solutions make contextual sense. Negative widths are impossible in reality, so recognizing \(x = 10\) as the logical solution is essential.

Each step is important to avoid errors and reach a practical conclusion, as seen with completing the word problem on the width of the deck.