Problem 82 - Algebra and Trigonometry

Question

Dimensions of a Room A rectangular bedroom is 7 \(\mathrm{ft}\) longer than it is wide. Its area is \(228 \mathrm{ft}^{2} .\) What is the width of the room?

Step-by-Step Solution

Answer

The width of the room is 12 feet.

Step 1: Understanding the Problem

We are given a rectangular bedroom where the length is 7 feet longer than the width. We know the area of the bedroom is 228 square feet, and need to find the width.

Step 2: Setting Up the Equation

Let the width of the room be \(w\) feet. Hence, the length of the room can be expressed as \(w + 7\) feet. The area of a rectangle is given by the formula \(\text{Area} = \text{length} \times \text{width}\). So we have the equation \(w(w + 7) = 228\).

Step 3: Expanding and Simplifying the Equation

Expand the equation by performing the multiplication:\[w^2 + 7w = 228\]Now we have a quadratic equation to solve for \(w\).

Step 4: Solving the Quadratic Equation

Rearrange the equation to standard form:\[w^2 + 7w - 228 = 0\]Now, we'll use the quadratic formula to solve for \(w\), where \(a = 1\), \(b = 7\), and \(c = -228\). The quadratic formula is:\[w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Step 5: Calculating the Solution

Substitute the values into the formula:\[w = \frac{-7 \pm \sqrt{7^2 - 4 \times 1 \times (-228)}}{2 \times 1}\]\[w = \frac{-7 \pm \sqrt{49 + 912}}{2}\]\[w = \frac{-7 \pm \sqrt{961}}{2}\]\[w = \frac{-7 \pm 31}{2}\]This gives us two potential solutions: \(w = \frac{24}{2} = 12\) and \(w = \frac{-38}{2} = -19\). The width cannot be negative, so \(w = 12\).

Step 6: Solution Verification

Check if the dimensions align with the area given. If the width is 12 feet, then the length is \(12 + 7 = 19\) feet. Calculating the area:\[\text{Area} = 12 \times 19 = 228\, \text{ft}^2\]This matches the given area, confirming our solution is correct.

Key Concepts

Rectangular Area Problems

Rectangular area problems are a classic way of applying basic geometry and algebra concepts to real-world situations. When dealing with rectangular objects or spaces, understanding how dimensions relate to area is crucial.

In our problem, the area is provided, and we need to solve for one of the dimensions. The area of a rectangle is calculated by multiplying its length by its width. This means if we can express the relationship between these measurements, we can set up an equation to find the unknown dimension.

In general, try to identify known information, such as one side being a certain amount longer or shorter than the other, and use these relationships to set up your formula.

Solving Quadratic Equations

Quadratic equations appear frequently in geometry problems like the ones involving areas of rectangles. A quadratic equation often takes the form:

• \[ax^2 + bx + c = 0\]

In many area problems, once the equation for the dimensions is set up, expanding and simplifying it results in a quadratic equation.

The quadratic equation is then solved using various methods, such as factoring (if possible), completing the square, or, as in this case, the quadratic formula:

• \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Recognizing this step is crucial since understanding how to manipulate and solve these expressions is essential to finding the needed dimension.

Algebraic Problem-Solving

Algebraic problem-solving involves using algebraic methods to find solutions to equations created from word problems. For these types of exercises, following a structured approach helps immensely, focusing on deriving and solving equations with accuracy.

The key steps involve:

• Identifying the variables and what they represent.
• Translating the word problem into an equation.
• Solving the equation using appropriate algebraic techniques.
• Verifying the solution makes sense in the context of the problem.

These steps transform rather confusing word problems into manageable algebraic challenges. Practicing these skills ensures one is well-prepared for more complex algebraic applications.

Geometry Applications

Geometry applications in real life often require determining unknown dimensions or areas for planning or design purposes. Understanding how geometric principles apply to problem-solving is key to addressing such real-world questions.

For instance, figuring out room dimensions is a common need in interior design, construction, or purchasing furniture. Knowing how to set up an equation from given measurements to find what you do not know is a valuable tool.

Real-life geometry not only involves solving such problems but also interpreting and double-checking solutions to ensure they are practical and valid.