Problem 346 - Elementary Algebra

Question

A rectangular retaining wall has area 15 square feet. The height of the wall is two feet less than its length. Find the height and the length of the wall.

Step-by-Step Solution

Answer

Length: 5 feet, Height: 3 feet.

Step 1: Define Variables

Let the length of the wall be denoted as \(L\) and the height of the wall be denoted as \(H\). Given that the height of the wall is two feet less than its length, we can express this relationship as: \(H = L - 2\).

Step 2: Area Formula Setup

The area of a rectangle is calculated as the product of its length and height. Therefore, the area equation is: \[ \text{Area} = L \times H = 15 \] Substitute \(H = L - 2\) into the area equation: \[ L \times (L - 2) = 15 \]

Step 3: Formulate Quadratic Equation

Expand the equation: \[ L^2 - 2L = 15 \] Then rearrange to form a standard quadratic equation: \[ L^2 - 2L - 15 = 0 \]

Step 4: Solve the Quadratic Equation

Factorize the quadratic equation: \[ (L - 5)(L + 3) = 0 \] Set each factor to zero and solve for \(L\): \[ L - 5 = 0 \implies L = 5 \ L + 3 = 0 \implies L = -3 \] Since length cannot be negative, we discard \(L = -3\), leaving us with \(L = 5\).

Step 5: Determine Height

Recall that \(H = L - 2\): \[ H = 5 - 2 = 3 \]

Step 6: Verify the Solution

Check the dimensions to confirm the solution: The area is \(5 \times 3 = 15\) square feet, which matches the problem statement.

Key Concepts

quadratic equations

Quadratic equations are polynomial equations of the second degree, usually expressed as \(ax^2 + bx + c = 0\). In our retaining wall problem, we ended up with the equation \(L^2 - 2L - 15 = 0\). This form allows us to solve for the variable, which in this case is the length of the wall.

To solve the quadratic equation, we can use factoring, completing the square, or the quadratic formula. Here, we factored the equation as follows:

• Factoring involves expressing the equation as a product of two binomials. For \(L^2 - 2L - 15\), we looked for two numbers that multiply to -15 and add up to -2. These numbers are -5 and 3.
• We then can write the equation as \((L - 5)(L + 3) = 0\).
• Setting each factor equal to zero gives us the possible solutions: \(L - 5 = 0\) or \(L + 3 = 0\). This simplifies to \(L = 5\) or \(L = -3\).

Since a negative length doesn't make sense in this context, we discard \(L = -3\). Therefore, the length \(L\) is 5 feet. Quadratic equations are essential in many geometry problems because they allow us to solve for unknown values based on given relationships.

rectangular area

When working with rectangular areas, it's important to understand how length and width (or height) interact. The area of a rectangle is given by the formula: \(\text{Area} = \text{Length} \times \text{Height}\). This formula tells us that the area is a result of multiplying the two dimensions of the rectangle.

In our retaining wall problem, we were given the area (15 square feet) and a relationship between the length and height: height is 2 feet less than the length. We can express this relationship as \(H = L - 2\).

Let’s see how we used this information:

• We replaced \(H\) in the area formula: \(L \times (L - 2) = 15\).
• This substitution turned our area formula into a quadratic equation when simplified and rearranged: \(L^2 - 2L - 15 = 0\).

Rectangular area problems often require you to create and solve equations based on one dimension being expressed in terms of another. Understanding the basic area formula and relationships between dimensions is crucial for solving such problems.

problem-solving

Effective problem-solving involves breaking down complex problems into smaller, manageable steps. Here's how we approached the retaining wall problem step by step:

• Define Variables: Start by identifying and defining what you're trying to find. Here, we defined \(L\) as the length and \(H\) as the height, with \(H = L - 2\).
• Set Up an Equation: Use the given information to create an equation. We used the area formula, \(L \times H = 15\), substituting \(H\) as \(L - 2\).
• Formulate and Solve the Quadratic Equation: Rearrange the equation into standard quadratic form and solve it through factoring or another method. We factored the equation to find \(L = 5\).
• Verify Solution: Always check your solution in the context of the problem to ensure it's correct. We verified that \(H = 3\) and recalculated the area to confirm it matched the given value.

Breaking down the problem this way makes it easier to tackle each part systematically. No step should be skipped as each contributes to the final solution. Remember, practice and repetition also help in honing your problem-solving skills.