Problem 72 - A Graphical Approach to Precalculus with Limits

Question

Solve each formula for the specified variable. \(d=\frac{1}{2} h\left(b_{1}+b_{2}\right)\) for \(b_{2} \quad\) (Area of a trapezoid)

Step-by-Step Solution

Answer

\(b_2 = \frac{2d}{h} - b_1\)

Step 1: Eliminate the fraction

The formula for the area of a trapezoid is given by \(d = \frac{1}{2} h (b_1 + b_2)\). To eliminate the fraction, multiply both sides of the equation by 2: \[ 2d = h (b_1 + b_2) \]

Step 2: Isolate the term with \(b_2\)

Next, to solve for \(b_2\), we need to get \((b_1 + b_2)\) on one side of the equation. We do this by dividing both sides of the equation by \(h\): \[ \frac{2d}{h} = b_1 + b_2 \]

Step 3: Solve for \(b_2\)

Now, isolate \(b_2\) by subtracting \(b_1\) from both sides of the equation: \[ b_2 = \frac{2d}{h} - b_1 \]

Key Concepts

Trapezoid Area Formula

Trapezoids are fascinating quadrilaterals with a pair of parallel sides. The formula to find the area of a trapezoid is quite unique. It calculates the space within those boundaries. Understanding this formula is crucial, especially if you're delving into precalculus topics. The formula is:

• \( d = \frac{1}{2} h (b_1 + b_2) \)

Here, \(d\) represents the area of the trapezoid, \(h\) is the height, and \(b_1\) and \(b_2\) are the lengths of the two parallel bases. The formula derives its value by taking the average of the two bases \((b_1 + b_2)/2\) and multiplying it by the height \(h\). This gives you an accurate measure of the trapezoid's area, enabling you to comprehend its spatial properties better. By applying this concept, students can tackle a variety of geometry problems involving quadrilaterals.

Solving Equations

Solving equations is all about finding the value of an unknown that satisfies the given relationship between numbers and variables. In the trapezoid problem, we were required to solve for \(b_2\). This means finding the specific value for \(b_2\) which makes the equation true. The process begins by simplifying the equation. If the equation includes fractions, a good first step is to eliminate those by multiplying through by the denominator. This makes the equation easier to work with. For example, multiplying the trapezoid's area formula by 2 canceled the \(\frac{1}{2}\), simplifying the equation significantly.This approach not only aids in solving for the desired variable but also enhances problem-solving skills across broader mathematical contexts.

Variable Isolation

Variable isolation is a crucial step in solving equations, especially when dealing with complex expressions. It involves isolating the variable of interest on one side of the equation. This makes it easier to understand and solve the problem. In our example, after eliminating the fraction, we worked towards isolating \(b_2\) by dividing both sides by \(h\), resulting in:

• \( \frac{2d}{h} = b_1 + b_2 \)

Continuing with this process, subtraction of \(b_1\) from both sides allowed us to completely solve for \(b_2\). Learning to isolate variables effectively is a skill that simplifies solving equations significantly. It decreases the complexity of mathematical expressions, making them more manageable. Practicing this skill equips students to deal with many types of equations in precalculus and other branches of mathematics.

Mathematical Expressions

Mathematical expressions form the backbone of equations and functions, containing numbers, variables, and operations. Understanding how to manipulate these expressions is crucial for solving algebraic problems. In the exercise given, the expression \( \frac{1}{2} h (b_1 + b_2) \) is a perfect example of where mathematical expressions are key. Simplifying this expression to solve for \(b_2\) required knowledge of arithmetic operations, properties of equality, and the distributive property. By rearranging and simplifying the original formula, we can derive new equations that help in finding solutions. Students need to be comfortable dissecting and rearranging these expressions. As they progress in precalculus, this skill will prove invaluable, helping them to easily switch between different forms of an equation or function. This understanding lays a strong foundation for tackling more advanced mathematical problems.