Problem 23 - Algebra 2 and Trigonometry

Question

In \(19-25,\) express each answer in simplest radical form. Check each answer. The lengths of the bases of a trapezoid are \(x+10\) and \(3 x+2\) and the length of the altitude is 2\(x\) . If the area of the trapezoid is \(40,\) find the lengths of the bases and of the altitude.

Step-by-Step Solution

Answer

The lengths are Base 1: 12, Base 2: 8, Altitude: 4.

Step 1: Understand the Problem

The problem involves finding the lengths of the bases and altitude of a trapezoid whose area is given. The lengths of the bases are expressions involving \(x\). The area of a trapezoid is found using the formula: \( \text{Area} = \frac{1}{2} \times \text{(base 1 + base 2)} \times \text{altitude} \).

Step 2: Set up the Equation

The area of the trapezoid is given as 40. Using the formula, we write the equation: \[ 40 = \frac{1}{2} \times ((x + 10) + (3x + 2)) \times 2x. \] Simplifying the right side, we combine like terms: \( (x + 10) + (3x + 2) = 4x + 12 \). Substitute this back into the equation.

Step 3: Simplify the Equation

Substitute back into the equation to get: \[ 40 = \frac{1}{2} \times (4x + 12) \times 2x. \] Simplify by multiplying the terms: \[ 40 = (4x + 12) \times x. \] Further simplification gives \[ 40 = 4x^2 + 12x. \]

Step 4: Solve the Quadratic Equation

We have the equation \( 4x^2 + 12x - 40 = 0 \). Divide the entire equation by 4 to simplify: \[ x^2 + 3x - 10 = 0. \] Factor the quadratic equation to find \(x\): \( (x + 5)(x - 2) = 0 \).

Step 5: Determine the Values of x

Solve each factor for zero: \( x + 5 = 0 \) gives \( x = -5 \) and \( x - 2 = 0 \) gives \( x = 2 \). Since negative lengths do not make sense in this context, \( x = 2 \) is the valid solution.

Step 6: Find Lengths of Bases and Altitude

Substitute \( x = 2 \) into the expressions for the bases and altitude. Base 1: \( x + 10 = 2 + 10 = 12 \). Base 2: \( 3x + 2 = 3(2) + 2 = 8 \). Altitude: \( 2x = 2(2) = 4 \).

Step 7: Check the Solution

Calculate the area with the found dimensions: \[ \text{Area} = \frac{1}{2} \times (12 + 8) \times 4 = \frac{1}{2} \times 20 \times 4 = 40. \] This confirms the solution, as the calculated area matches the given area.

Key Concepts

Quadratic Equation

A quadratic equation is a type of polynomial equation of the second degree, which means its highest exponent on the variable is 2. It generally appears in the standard form: \( ax^2 + bx + c = 0 \). In our trapezoid problem, we arrived at a quadratic equation when simplifying the expressions for the area of the trapezoid. The equation to solve was \( 4x^2 + 12x - 40 = 0 \). To make solving this equation easier, it can be simplified by dividing through by 4, giving us \( x^2 + 3x - 10 = 0 \). This converts the quadratic into a manageable form that can be factored. When solving quadratic equations, there are several methods:

• Factoring
• Completing the square
• Using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

In our case, factoring was straightforward because the quadratic neatly broke into \( (x + 5)(x - 2) = 0 \). Solving each factor yields the potential solutions for \( x \). It's essential to consider both solutions and contextualize them with the problem, as negative solutions often don't make sense for geometric problems involving lengths.

Simplest Radical Form

Expressing numbers in their simplest radical form involves writing roots, like square roots, as simplified as possible. This means breaking down the radicand (the number under the root) into its prime factors and simplifying it by pulling out pairs from under the root where possible.For instance, the number 40 can be written as \( 5 \times 8 \). To simplify \( \sqrt{40} \), we first break down 8 into \( 2 \times 4 \), and 4 into \( 2 \times 2 \). So \( \sqrt{40} = \sqrt{5 \times 2 \times 2 \times 2} \). This can be further simplified as \( \sqrt{4 \times 10} = 2\sqrt{10} \). Therefore, \( 2\sqrt{10} \) is the simplest radical form. In the context of our trapezoid problem, solving for \(x\) doesn't directly provide a scenario for calculating a radical form. However, understanding radical simplification is useful if encountering non-integer lengths expressed under square roots.

Base and Altitude Length

In geometry, especially when dealing with trapezoids, it is crucial to understand what is meant by the "base" and "altitude." A trapezoid is a four-sided flat shape with one pair of parallel sides, known as the bases. The altitude is the perpendicular distance between these two bases.When given expressions for the bases and altitude as we had \((x + 10)\), \((3x + 2)\), and \(2x\) respectively, finding their specific lengths involves plugging in a valid solution for \(x\). In this exercise, we solved the quadratic equation and found \(x = 2\). Thus:

• Base 1: \( x + 10 = 12 \)
• Base 2: \( 3x + 2 = 8 \)
• Altitude: \( 2x = 4 \)

Understanding the relationship between the bases and altitude is crucial when using the area formula: \[ \text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Altitude} \] This relationship allows us to solve for unknowns when other dimensions are provided, making it a fundamental part of tackling similar trapezoidal problems.