Problem 38 - Elementary Algebra

Question

The height of a triangle is 4 units less than the length of the base. If the area of the triangle is 48 square units, then find the length of its base and height.

Step-by-Step Solution

Answer

The base is 12 units and the height is 8 units.

Step 1: Identify the Given Information

We know that the area of the triangle is 48 square units and the height is 4 units less than the base. We need to find both the base and the height of the triangle.

Step 2: Set Up the Equation for Rectangle Area

The formula for the area of a triangle is \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \). Let the base be \( b \) and the height \( h = b - 4 \). We can set up the equation as follows: \( \frac{1}{2} \times b \times (b - 4) = 48 \).

Step 3: Simplify and Solve the Quadratic Equation

First, multiply both sides by 2 to eliminate the fraction: \( b(b - 4) = 96 \). Expand the equation to \( b^2 - 4b = 96 \). Rearrange it to form a standard quadratic equation: \( b^2 - 4b - 96 = 0 \).

Step 4: Apply the Quadratic Formula

Use the quadratic formula \( b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \) where \( A = 1 \), \( B = -4 \), and \( C = -96 \).Calculate the discriminant: \( B^2 - 4AC = (-4)^2 - 4 \cdot 1 \cdot (-96) = 16 + 384 = 400 \).Compute the solutions: \( b = \frac{-(-4) \pm \sqrt{400}}{2 \cdot 1} \) which gives \( b = \frac{4 \pm 20}{2} \).

Step 5: Find Base and Height

Calculate both roots of the quadratic equation: \( b = \frac{24}{2} = 12 \) and \( b = \frac{-16}{2} = -8 \). Since a negative base length is not possible, the base \( b \) is 12 units. Substituting back, the height \( h = b - 4 = 12 - 4 = 8 \) units.

Key Concepts

Solving Quadratic Equations

Quadratic equations form a crucial component of algebra, involving equations of the form \( ax^2 + bx + c = 0 \). These equations may have two solutions, which are calculated using the quadratic formula:

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

In our problem, the quadratic equation was formed when we derived the equation for the triangular area as \( b(b - 4) = 96 \). Upon rearranging, it became \( b^2 - 4b - 96 = 0 \). Here, we have:

• \( a = 1 \)
• \( b = -4 \)
• \( c = -96 \)

To solve, the discriminant - \( b^2 - 4ac \) - plays a vital role. A positive discriminant, as we found (\(400\)), indicates two real solutions. By applying the formula, we calculated both possible values for \( b \), ultimately selecting the one that fits the physical context of the problem.

Triangle Area Formula

Triangles have a unique formula for area calculation: \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \). This formula helps determine unknown dimensions when the area is provided. In the exercise, the base was labeled as \( b \) and the height as \( h = b - 4 \), leading to the equation \( \frac{1}{2} \times b \times (b - 4) = 48 \).
To remove the fraction and streamline solving, both equation sides were multiplied by 2, yielding \( b(b - 4) = 96 \). Problem-solving often involves creating a manageable equation from the given formula. Hence, understanding how to manipulate this formula allows students to approach diverse triangular problems confidently.

Algebraic Problem Solving

Algebraic problem-solving integrates concepts like forming equations, manipulating algebraic expressions, and solving them, as demonstrated in this problem. The goal is to establish a link between unknowns using known quantities.

• Express the known relationships, like height being 4 units less than the base.
• Formulate a target equation using these relations and solve it to find unknowns.

This analytical process involves translating a real-world problem's scenario into mathematical expressions, followed by solving the equations formed. Here, after setting \( h = b - 4 \), the derived equation \( b^2 - 4b - 96 = 0 \) was tackled using quadratic equation-solving techniques. Through logical steps and systematic calculations, students learn to break down complex problems into simple, solvable parts.