Problem 36
Question
A parallelogram with an area of \(6 x^{2}-7 x-5\) square units has a base of \(3 x-5\) units. Determine the height of the parallelogram.
Step-by-Step Solution
Verified Answer
The height of the parallelogram is \(2x + 1\) units.
1Step 1: Understand the Formula for Area of a Parallelogram
The area of a parallelogram can be calculated using the formula \( \text{Area} = \text{base} \times \text{height} \). In this problem, we are given the area as \(6x^2 - 7x - 5\) and the base as \(3x - 5\). We need to find the height \(h\).
2Step 2: Substitute Known Values into the Formula
Substitute the given expressions into the formula:\[6x^2 - 7x - 5 = (3x - 5) \times h.\]
3Step 3: Solve for the Height
To find the height \(h\), solve the equation by isolating \(h\):\[h = \frac{6x^2 - 7x - 5}{3x - 5}.\]
4Step 4: Simplify the Expression
To ensure the expression is fully simplified, use polynomial division to divide \(6x^2 - 7x - 5\) by \(3x - 5\). After performing the division, we find that:\[h = 2x + 1.\] (Ensure to double-check the division and any remainder is managed properly.)
Key Concepts
Parallelogram Area FormulaPolynomial DivisionAlgebraic Expressions
Parallelogram Area Formula
The parallelogram area formula is essential in geometry, allowing you to calculate the area of a parallelogram using its base and height. The formula itself is straightforward:
In practical problems, such as our exercise, this formula is used inversely. We start with the area and one dimension (the base) to determine the unknown dimension (the height). Recognizing and understanding how to manipulate this formula in different scenarios is crucial for solving real-world problems.
- Area = Base × Height
In practical problems, such as our exercise, this formula is used inversely. We start with the area and one dimension (the base) to determine the unknown dimension (the height). Recognizing and understanding how to manipulate this formula in different scenarios is crucial for solving real-world problems.
Polynomial Division
Polynomial division is a key algebraic tool which is similar to long division conducted with numbers. Here, you take a polynomial (dividend) and divide it by another polynomial (divisor), simplifying the expression across the division.
The process involves these critical steps:
This demonstrates how polynomial division helps simplify complex expressions into more digestible parts.
The process involves these critical steps:
- Identify the leading term in both the dividend and the divisor.
- Determine how many times the leading term of the divisor goes into the leading term of the dividend.
- Multiply this result by the entire divisor and subtract this from the original dividend.
- Repeat the process with the new dividend until the remainder is smaller than the leading term of the divisor.
This demonstrates how polynomial division helps simplify complex expressions into more digestible parts.
Algebraic Expressions
Algebraic expressions are mathematical phrases that can contain numbers, variables, and operational symbols. They are the fundamental building blocks of algebra and can represent various forms of relationships and equations.
Understanding algebraic expressions involves recognizing components like coefficients, variables, and constants. For example, in the expression \(6x^2 - 7x - 5\),
By transforming complex expressions into simpler ones, using techniques like the polynomial division discussed earlier, you effectively apply algebraic skills to find solutions efficiently.
Understanding algebraic expressions involves recognizing components like coefficients, variables, and constants. For example, in the expression \(6x^2 - 7x - 5\),
- \(6x^2\) represents a term with 6 as a coefficient and \(x^2\) as the variable part.
- \(-7x\) is another term with \(-7\) as the coefficient and \(x\) as the variable.
- \(-5\) is a constant term.
By transforming complex expressions into simpler ones, using techniques like the polynomial division discussed earlier, you effectively apply algebraic skills to find solutions efficiently.
Other exercises in this chapter
Problem 36
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