Problem 22 - Algebra 2 and Trigonometry

Question

The length of the base of a triangle is 6 feet more than the length of the altitude to the base. The area of the triangle is 18 square feet. Find the length of the base and of the altitude to the base.

Step-by-Step Solution

Answer

The altitude is approximately 3.7 feet, and the base is approximately 9.7 feet.

Step 1: Define Variables

Let's denote the length of the altitude of the triangle as \( x \) feet. Since the base is 6 feet more than the altitude, the base can be expressed as \( x + 6 \) feet.

Step 2: Area Formula for a Triangle

The formula for the area of a triangle is \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{altitude} \). Substituting the given area, base, and altitude values into the formula gives: \( 18 = \frac{1}{2} \times (x + 6) \times x \).

Step 3: Set Up the Equation

Simplify the equation: \( 18 = \frac{1}{2} \times (x^2 + 6x) \). Multiply both sides by 2 to clear the fraction: \( 36 = x^2 + 6x \).

Step 4: Solve the Quadratic Equation

Rearrange the equation: \( x^2 + 6x - 36 = 0 \). Use the quadratic formula: \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \), where \( a = 1 \), \( b = 6 \), and \( c = -36 \).

Step 5: Calculate the Discriminant

Find the discriminant \( b^2 - 4ac = 6^2 - 4 \times 1 \times (-36) = 36 + 144 = 180 \).

Step 6: Find the Roots Using Quadratic Formula

Calculate the roots: \( x = \frac{{-6 \pm \sqrt{180}}}{2} \). Simplify further: \( x = \frac{{-6 \pm 6\sqrt{5}}}{2} \), which gives \( x = -3 \pm 3\sqrt{5} \). Since \( x \) represents length, choose the positive root: \( x = -3 + 3\sqrt{5} \).

Step 7: Calculate the Base and Altitude

The altitude is \( x = -3 + 3\sqrt{5} \) feet, and the base is \( x + 6 = 3\sqrt{5} + 3 \) feet.

Key Concepts

Triangle Area

The area of a triangle is a measure of the space enclosed within its three sides. It's a basic but important concept in geometry. To find the area, you can use the formula:

• \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{altitude}\)

The base is any one of the three sides of the triangle, and the altitude (or height) is the perpendicular line segment from the base to the opposite vertex.
This formula works because it essentially treats the triangle as half of a rectangle, which has an area of base times height.
So, when solving for a triangle's area, it's crucial to correctly identify these dimensions in order to compute accurately.

Variable Definition

In algebra and mathematics, defining variables helps in setting up equations to solve problems. Variables act as placeholders for unknown values, making complex problems more manageable.
Here, we defined:

• \( x \) as the length of the altitude of the triangle,
• \( x + 6 \) as the length of the base, since it is 6 feet longer than the altitude.

By establishing these variables, we can easily substitute them back into equations that stem from important geometric principles and relationships. This step not only simplifies the process but also ensures clarity and consistency throughout the calculations.
Understanding how to define variables effectively is a foundational skill in algebra and problem-solving.

Discriminant

The discriminant is a critical part of the quadratic formula. It indicates the nature of the roots of a quadratic equation. The discriminant (\(b^2 - 4ac\)) is derived from the quadratic equation in its standard form \(ax^2 + bx + c = 0\).
Depending on the value of the discriminant:

• If it's positive, two distinct real roots exist.
• If it's zero, there's exactly one real root (also called a repeated root).
• If it's negative, no real roots exist, only complex numbers.

In our exercise, the discriminant value was 180, a positive number, indicating two possible real roots. This helped us determine the proper solution for the triangle dimensions.

Quadratic Formula

The quadratic formula is used to find the roots of a quadratic equation. The formula is:

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

This formula lets you solve any quadratic equation by plugging the coefficients \(a\), \(b\), and \(c\) into it.
In this case, the equation was \(x^2 + 6x - 36 = 0\), so: \(a = 1\), \(b = 6\), and \(c = -36\). Using the quadratic formula, we computed the possible lengths for \(x\).
Understanding the quadratic formula is key for solving complex polynomial equations and is widely used in various fields of science and engineering.