Problem 93
Question
Solve for the indicated variable. Area of a Trapezoid Solve for \(b\) in \(A=\frac{1}{2}(a+b) h\).
Step-by-Step Solution
Verified Answer
The solution for \(b\) in the given trapezoid area formula is \(b=\frac{2A - ah}{h}\).
1Step 1: Understand the question and the given formula
The question requires solving for the variable \(b\), given the formula for the area of a trapezoid \(A=\frac{1}{2}(a+b)h\).
2Step 2: Remove fraction
To remove the fraction, we can multiply the full equation by 2. This gives \(2A= (a+b)h\).
3Step 3: Isolate terms with the variable b
Isolate the terms with \(b\) on one side of the equation by subtracting \(ah\) from both sides. This gives \(2A - ah = bh\).
4Step 4: Solve for b
Next, divide both sides by \(h\) to solve for \(b\). This gives \(b=\frac{2A - ah}{h}\). This is the final solution.
Key Concepts
Area of a Trapezoid FormulaAlgebraic ManipulationIsolating Variables
Area of a Trapezoid Formula
Understanding the area of a trapezoid formula is crucial before one can delve into solving for any of its variables. A trapezoid is a four-sided figure with one pair of parallel sides. The formula to calculate its area is given as:\[ A = \frac{1}{2}(a+b)h \]
Here, \( A \) represents the area of the trapezoid, \( a \) and \( b \) are the lengths of the two parallel sides, often referred to as the bases, and \( h \) denotes the height, which is the perpendicular distance between these bases.
Visualizing this formula as the area of two right triangles and a rectangle combined can help students see why it works. This visualization approach ties into the exercise improvement advice of making mathematical concepts accessible through real-world examples and visual aids.
Here, \( A \) represents the area of the trapezoid, \( a \) and \( b \) are the lengths of the two parallel sides, often referred to as the bases, and \( h \) denotes the height, which is the perpendicular distance between these bases.
Visualizing this formula as the area of two right triangles and a rectangle combined can help students see why it works. This visualization approach ties into the exercise improvement advice of making mathematical concepts accessible through real-world examples and visual aids.
Algebraic Manipulation
Algebraic manipulation involves rearranging terms, factoring equations, and performing operations to simplify expressions or solve for variables. In the context of our example, removing fractions is a common step in algebraic manipulation. By multiplying the area formula by 2, we eliminate the fraction and simplify the equation. Understanding how to manipulate algebraic expressions is vital because it allows us to change the form of an equation to make other operations, like solving for a variable, more straightforward.
Using proper algebraic manipulation techniques will ensure that the integrity of the equation remains intact while making progress towards the solution. This step adheres to the suggested improvement of ensuring a clear progression from one step to the next without skipping any logical, arithmetic, or algebraic steps.
Using proper algebraic manipulation techniques will ensure that the integrity of the equation remains intact while making progress towards the solution. This step adheres to the suggested improvement of ensuring a clear progression from one step to the next without skipping any logical, arithmetic, or algebraic steps.
Isolating Variables
Isolating variables is a fundamental skill in algebra that involves getting the variable of interest alone on one side of the equation to solve for its value. This process often involves operations opposite to those already performed. For solving for \( b \) in our trapezoid area problem, we needed to move all terms not containing \( b \) to the other side of the equation. By subtracting \( ah \) from both sides, we successfully isolated terms that contained \( b \).
Once isolated, we can then divide by the coefficient of \( b \) to solve for it directly. The coherent use of this technique is what allows students to transition from complex equations to simple, solvable forms. It embodies the advice from the textbook solution improvement to guide students through the process of breaking down and simplifying each individual algebraic step.
Once isolated, we can then divide by the coefficient of \( b \) to solve for it directly. The coherent use of this technique is what allows students to transition from complex equations to simple, solvable forms. It embodies the advice from the textbook solution improvement to guide students through the process of breaking down and simplifying each individual algebraic step.
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Problem 93
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