Problem 59 - Elementary Algebra

Question

The area of a triangle is 14 square feet. If the base is 4 feet more than 2 times the height, then find the length of the base and the height.

Step-by-Step Solution

Answer

The height is 2 feet, and the base is 8 feet.

Step 1: Understand the Formula for the Area of a Triangle

The area of a triangle can be calculated using the formula \( A = \frac{1}{2} imes \text{base} \times \text{height} \). In this problem, the area is given as 14 square feet.

Step 2: Set Up the Relationship Between Base and Height

We are given that the base is 4 feet more than 2 times the height. If we let \( h \) represent the height, then the base can be expressed as \( b = 2h + 4 \).

Step 3: Substitute the Expressions into the Area Formula

Substitute \( b = 2h + 4 \) and \( A = 14 \) into the area formula: \( 14 = \frac{1}{2} \times (2h + 4) \times h \).

Step 4: Simplify and Solve for Height

First, eliminate the fraction by multiplying both sides by 2: \( 28 = (2h + 4) \times h \). Expand and move terms: \( 28 = 2h^2 + 4h \). Rearrange to form a quadratic equation: \( 2h^2 + 4h - 28 = 0 \).

Step 5: Factor the Quadratic Equation

Divide the entire equation by 2 to simplify: \( h^2 + 2h - 14 = 0 \). This can be factored to \((h + 7)(h - 2) = 0\).

Step 6: Solve for Possible Values of Height

From the factored form, solve for \( h \): \( h + 7 = 0 \) or \( h - 2 = 0 \). Hence, \( h = -7 \) or \( h = 2 \). Since height cannot be negative, we have \( h = 2 \).

Step 7: Find the Length of the Base

Use the expression for the base \( b = 2h + 4 \) with \( h = 2 \): \( b = 2(2) + 4 = 4 + 4 = 8 \). Thus, the base is 8 feet.

Key Concepts

Understanding Quadratic Equations

Quadratic equations are polynomial equations of degree two. They take the form \( ax^2 + bx + c = 0 \), where "a," "b," and "c" are constants, and \( x \) represents the variable. These equations are ubiquitous in algebra and can describe a broad set of problems in both theoretical and applied contexts, such as physics, engineering, and geometry.
In the triangle area problem we just tackled, we derived a quadratic equation when converting our geometric expression into an algebraic one: \( 2h^2 + 4h - 28 = 0 \).
This transformation occurs when expressing one unknown variable (base, in terms of height) and substituting it back into a formula to solve for another variable (height).
Quadratic equations can be solved using various methods:

• Factoring: Expressing the quadratic as a product of two binomials.
• Quadratic Formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), which is a universal method.
• Completing the Square: Rewriting the equation to make it easier to solve.

Identifying quadratic expressions in geometric contexts often involves representing shapes or dimensions through algebraic terms. This is beneficial because it allows us to leverage algebraic tools to solve geometric problems.

Factoring the Quadratic Equation

Factoring is a powerful algebraic technique used to solve quadratic equations by expressing them as the product of their binomial roots. The goal in factoring is to write the equation \( ax^2 + bx + c = 0 \) as \((mx + n)(px + q) = 0\).
This method relies on finding two numbers that both add up to the coefficient of the middle term \( b \) and multiply to the product of the leading coefficient \( a \) and the constant term \( c \).
In our original problem, after simplifying the equation into \( h^2 + 2h - 14 = 0 \), the task was to factor this quadratic equation. Here, it simplifies neatly to \((h + 7)(h - 2) = 0\).
By setting each binomial to zero, we find the possible values for \( h \):

• \( h + 7 = 0 \) leads to \( h = -7 \).
• \( h - 2 = 0 \) leads to \( h = 2 \).

Since a negative height isn’t feasible in geometry, \( h = 2 \) is our solution. Factoring, therefore, transformed the quadratic challenge into a simple process of elimination, pinpointing the realistic solution.

Using Geometry Formulas in Problem Solving

Geometry formulas are a fundamental part of solving spatial problems, especially when identifying areas, perimeters, and volumes. These formulas allow us to calculate these quantities quickly when given specific dimensions.
The area of a triangle can be efficiently found using the formula \( A = \frac{1}{2} \times \text{base} \times \text{height} \). In our problem, this formula was crucial for setting up the initial equation.
We were provided with a relationship: the base was given as "4 feet more than 2 times the height," expressed algebraically as \( b = 2h + 4 \).
After substituting this into the area formula \( A = \frac{1}{2} \times (2h + 4) \times h \), it let us derive the quadratic equation used to find the height. Solving this helped us find the height, which was then used to easily determine the base length.
Key points to remember about geometry formulas:

• Memory aids: They reduce the complexity of manual calculations.
• Flexibility: Many can be adapted or expanded based on contextual modifications, like expressing known relationships.
• Wide application: Found in various geometric shapes, from polygons to circles and beyond.

Thoroughly understanding these formulas enhances your ability to solve diverse mathematical challenges effectively.