Problem 81 - Algebra and Trigonometry

Question

Dimensions of a Garden A rectangular garden is 10 \(\mathrm{ft}\) longer than it is wide. Its area is \(875 \mathrm{ft}^{2} .\) What are its dimensions?

Step-by-Step Solution

Answer

The garden is 25 ft wide and 35 ft long.

Step 1: Define Variables

Let's define the variable for the width of the garden. Let \( w \) represent the width of the garden in feet.

Step 2: Express the Length in Terms of the Width

According to the problem, the length of the garden is 10 feet longer than its width. Therefore, we can express the length \( l \) as \( l = w + 10 \).

Step 3: Write the Equation for the Area

The area of the garden is given by the formula for the area of a rectangle, \( A = l \times w \). Plug in the expressions for area, width, and length: \( 875 = (w + 10) \times w \).

Step 4: Simplify and Rearrange the Equation

Expand the equation from step 3: \( 875 = w^2 + 10w \). Rearrange to form a quadratic equation: \( w^2 + 10w - 875 = 0 \).

Step 5: Solve the Quadratic Equation

To solve \( w^2 + 10w - 875 = 0 \), use the quadratic formula: \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 10 \), \( c = -875 \). Calculate the discriminant: \( 10^2 - 4\times 1 \times (-875) = 100 + 3500 = 3600 \). So, \( w = \frac{-10 \pm \sqrt{3600}}{2} \). \( \sqrt{3600} = 60 \), thus \( w = \frac{-10 \pm 60}{2} \). This gives two solutions: \( w = 25 \) or \( w = -35 \). Since width cannot be negative, \( w = 25 \).

Step 6: Find the Corresponding Length

Now that we know \( w = 25 \), plug it back into the expression for length: \( l = w + 10 \). So, \( l = 25 + 10 = 35 \).

Step 7: Verify the Solution

Verify the solution by calculating the area with the found dimensions: \( A = l \times w = 35 \times 25 = 875 \mathrm{ft}^2 \), which matches the problem statement.

Key Concepts

Rectangular Garden Problems

Rectangular garden problems are common in geometry and algebra because they help us understand how to work with shapes and equations. In such problems, we typically need to find dimensions given area and other relationships. Let's say we have a rectangular garden where we know the length is a certain amount longer than the width, and we have the total area. This is a typical setup for a rectangular garden problem:

• We define a variable for one of the dimensions, usually the width or length.
• We express the other dimension in terms of this variable using the given relationship (e.g., the length is 10 feet longer than the width).
• Then, we use the formula for the area of a rectangle to set up an equation.

These problems often require logical thinking and the ability to create and solve algebraic equations, making them a great exercise for critical thinking skills.

Area of Rectangles

The area of a rectangle is a fundamental concept in geometry. The formula to find the area of a rectangle is simple: multiply the length by the width. This is because a rectangle has two pairs of equal opposite sides, and its area is essentially the total amount of space enclosed within its sides. Here's the basic formula:

• Area, \( A = l \times w \)

where \( l \) is the length, and \( w \) is the width of the rectangle.Understanding the area helps in many practical scenarios, such as calculating the floor space in a room, planning a garden's layout, or anything involving space measurement. When working with word problems like this, identifying these measurements and applying the area formula correctly is crucial for finding the right solution.

Solving Quadratic Equations

Quadratic equations form the backbone of many algebra problems, including those related to geometry. A quadratic equation typically has the form \( ax^2 + bx + c = 0 \). Solving these equations is crucial when dealing with polynomial relationships, like in our rectangular garden problem. To solve the quadratic equation \( w^2 + 10w - 875 = 0 \), we can use the quadratic formula:

• \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

where:

• \( a = 1 \),
• \( b = 10 \),
• \( c = -875 \).

The quadratic formula provides two potential solutions, but in practical word problems, like the dimensions of a garden, we focus on realistic values (e.g., positive dimensions). Understanding how to apply this formula is essential to navigate problems that go beyond basic linear equations. Breaking down each step clearly and validating results with the context of the problem is vital to ensure accurate solutions.