Problem 45 - Elementary Algebra Within Reach

Question

Structure Write three equations that are equivalent to \(A=\frac{1}{2}(x+y) h\) by solving for each variable, where \(A\) is the area, \(h\) is the height, and \(x\) and \(y\) are the bases of a trapezoid. Explain when you would use each equation.

Step-by-Step Solution

Answer

The three equivalent equations are \(x = \frac{2A}{h} - y\), \(y = \frac{2A}{h} - x\), and \(h = \frac{2A}{x+y}\). The choice of equation to use depends on which quantities are known and which quantity one wants to find.

Step 1: Solve for x

First, solve for \(x\). Starting from the given equation \(A=\frac{1}{2}(x+y)h\), isolate \(x\) by subtracting \(y\) from both sides and then multiplying both sides by \(2/h\). The resultant equation is \(x = \frac{2A}{h} - y\) which can be used when the area \(A\), the height \(h\), and the length of one base \(y\) are known and one wants to find the length of the other base \(x\).

Step 2: Solve for y

Next, solve for \(y\). This is a similar process to solving for \(x\). Subtract \(x\) from both sides and then multiply both sides by \(2/h\) to get the equation \(y = \frac{2A}{h} - x\). This equation can be used when the area \(A\), the height \(h\), and the length of one base \(x\) are known and one wants to find the length of the other base \(y\).

Step 3: Solve for h

Finally, solve for \(h\). To isolate \(h\), divide both sides by \((x+y)/2\). The resultant equation is \(h = \frac{2A}{x+y}\). This equation can be used when the area \(A\) and the lengths of the two bases \(x\) and \(y\) are known and one wants to find the height \(h\).

Key Concepts

Trapezoid Area Formula

When it comes to understanding shapes, the trapezoid holds a unique place with its pair of parallel sides known as 'bases'. Calculating the area of this quadrilateral requires a special formula, which is:
\[ A = \frac{1}{2}(x+y)h \]
In this equation, \( A \) denotes the area of the trapezoid, \( x \) and \( y \) represent the lengths of the two bases, and \( h \) stands for the height, which is the perpendicular distance between the bases.
Learning to work with this formula is not just about plugging values into variables; it is a doorway to understanding the properties of trapezoids and how they relate to each other. This equation can be rearranged to solve for various elements, which is useful in different scenarios. For example, a construction worker might use it to calculate the area of a trapezoidal concrete block, while a tailor could apply it to designing a piece of fabric with a trapezoidal shape.

Isolating Variables

Key to solving equations in algebra is the ability to isolate variables—it's like getting one participant to speak up in a crowded, noisy room. When we isolate a variable, we are essentially rearranging the equation so that this variable stands alone on one side, providing its value in terms of the other known quantities.
Isolating variables is a fundamental skill in algebra that allows you to

• Understand the relationship between different elements in an equation.
• Solve for unknowns, facilitating the process of finding answers.
• Make predictions by setting up equations based on given data.

To isolate a variable, we perform operations such as addition, subtraction, multiplication, or division, and we apply these operations equally on both sides of the equation to maintain its balance. For instance, in the equation
\[ x = \frac{2A}{h} - y \]
the variable \( x \) is isolated. We reached this form by strategically moving other variables and constants around. Being comfortable with this process is crucial for students who wish to excel in algebra and beyond.

Algebraic Manipulation

Algebraic manipulation, also known as algebraic operations, is essentially the toolkit for working with equations and formulas. It involves a set of techniques to rearrange, simplify, or solve algebraic expressions and equations.
It includes a variety of moves you might make, such as:

• Adding or subtracting terms on both sides of an equation.
• Multiplying or dividing both sides by a number (or another expression).
• Expanding expressions using the distributive property.
• Factoring expressions to simplify them or to solve equations.
• Combining like terms to condense expressions.

Through these methods, we can redesign complex equations into simpler forms, uncovering the value of unknowns in the process. The goal is to become so fluent with these operations that, much like a chess player thinks several moves ahead, a student can look at a complex equation and strategize the steps to isolate each variable, as seen in the steps to solve for \( x \), \( y \), and \( h \) in the trapezoid area equation. This proficiency is valuable not just in completing math homework, but in applying mathematical reasoning in real-world situations.