Problems Involving Formulas II
Learning Outcomes
- Solve a formula for a specific variables
- Evaluate a formula using substitution
- Solve area and perimeter words problems
- Solve temperature conversion problems
Geometry
There are many geometric shapes that have been well studied over the years. We know quite a bit about circles, rectangles, and triangles. Mathematicians have proven many formulas that describe the dimensions of geometric shapes including area, perimeter, surface area, and volume.Perimeter
Perimeter is the distance around an object. For example, consider a rectangle with a length of [latex]8[/latex] and a width of [latex]3[/latex]. There are two lengths and two widths in a rectangle (opposite sides), so we add [latex]8+8+3+3=22[/latex]. Since there are two lengths and two widths in a rectangle, you may find the perimeter of a rectangle using the formula [latex]{P}=2\left({L}\right)+2\left({W}\right)[/latex] where [latex]L[/latex] = Length [latex]W[/latex] = Width In the following example, we will use the problem-solving method we developed to find an unknown width using the formula for the perimeter of a rectangle. By substituting the dimensions we know into the formula, we will be able to isolate the unknown width and find our solution.Example
You want to make another garden box the same size as the one you already have. You write down the dimensions of the box and go to the lumber store to buy some boards. When you get there, you realize you didn't write down the width dimension—only the perimeter and length. You want the exact dimensions so you can have the store cut the lumber for you. Here is what you have written down: Perimeter = [latex]16.4[/latex] feet Length = [latex]4.7[/latex] feet Can you find the dimensions you need to have your boards cut at the lumber store? If so, how many boards do you need and what lengths should they be?Answer: Read and Understand: We know perimeter = [latex]16.4[/latex] feet and length = [latex]4.7[/latex] feet, and we want to find width. Define and Translate: Define the known and unknown dimensions: [latex]W[/latex] = width [latex-display]P[/latex] = [latex]16.4[/latex-display] [latex-display]L[/latex] = [latex]4.7[/latex-display] Write and Solve: First, we will substitute the dimensions we know into the formula for perimeter:
[latex]\begin{array}{L}\,\,\,\,\,P=2{W}+2{L}\\\\16.4=2\left(W\right)+2\left(4.7\right)\end{array}[/latex]
Then, we will isolate [latex]W[/latex] to find the unknown width.[latex]\begin{array}{L}16.4=2\left(W\right)+2\left(4.7\right)\\16.4=2{W}+9.4\\\underline{-9.4\,\,\,\,\,\,\,\,\,\,\,\,\,-9.4}\\\,\,\,\,\,\,\,7=2\left(W\right)\\\,\,\,\,\,\,\,\frac{7}{2}=\frac{2\left(W\right)}{2}\\\,\,\,\,3.5=W\end{array}[/latex]
Write the width as a decimal to make cutting the boards easier and replace the units on the measurement, or you won't get the right size of board! Check and Interpret: If we replace the width we found, [latex]W=3.5\text{ feet }[/latex] into the formula for perimeter with the dimensions we wrote down, we can check our work:[latex]\begin{array}{L}\,\,\,\,\,{P}=2\left({L}\right)+2\left({W}\right)\\\\{16.4}=2\left({4.7}\right)+2\left({3.5}\right)\\\\{16.4}=9.4+7\\\\{16.4}=16.4\end{array}[/latex]
Our calculation for width checks out. We need to ask for [latex]2[/latex] boards cut to [latex]3.5[/latex] feet and [latex]2[/latex] boards cut to [latex]4.7[/latex] feet so we can make the new garden box.Example
Isolate the term containing the variable, [latex]W[/latex], from the formula for the perimeter of a rectangle:[latex]{P}=2\left({L}\right)+2\left({W}\right)[/latex].
Answer: First, isolate the term with [latex]W[/latex] by subtracting [latex]2L[/latex] from both sides of the equation.
[latex] \displaystyle \begin{array}{L}\,\,\,\,\,\,\,\,\,\,P\,=\,\,\,\,2L+2W\\\underline{\,\,\,\,\,-2L\,\,\,\,\,-2L\,\,\,\,\,\,\,\,\,\,\,}\\P-2L=\,\,\,\,\,\,\,\,\,\,\,\,\,2W\end{array}[/latex]
Next, clear the coefficient of [latex]W[/latex] by dividing both sides of the equation by [latex]2[/latex].[latex]\displaystyle \begin{array}{L}\underline{P-2L}=\underline{2W}\\\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\\ \,\,\,\frac{P-2L}{2}\,\,=\,\,W\end{array}[/latex]
You can rewrite the equation so the isolated variable is on the left side.[latex]W=\frac{P-2L}{2}[/latex]
Area
The area of a triangle is given by [latex] A=\frac{1}{2}bh[/latex] where [latex]A[/latex] = area [latex]b[/latex] = the length of the base [latex]h[/latex] = the height of the triangle Remember, that when two variables, or a number and a variable, are sitting next to each other without a mathematical operator between them, you can assume they are being multiplied. This can seem frustrating, but you can think of it like mathematical slang. Over the years, people who use math frequently have just made that shortcut enough times that it has been adopted as convention. In the next example, we will use the formula for area of a triangle to find a missing dimension, as well as use substitution to solve for the base of a triangle given the area and height.Example
Find the base ([latex]b[/latex]) of a triangle with an area ([latex]A[/latex]) of [latex]20[/latex] square feet and a height ([latex]h[/latex]) of [latex]8[/latex] feet.Answer: Use the formula for the area of a triangle, [latex] {A}=\frac{{1}}{{2}}{bh}[/latex]. Substitute the given lengths into the formula and solve for [latex]b[/latex].
[latex]\displaystyle \begin{array}{l}\,\,A=\frac{1}{2}bh\\\\20=\frac{1}{2}b\cdot 8\\\\20=\frac{8}{2}b\\\\20=4b\\\\\frac{20}{4}=\frac{4b}{4}\\\\ \,\,\,5=b\end{array}[/latex]
Answer
The base of the triangle measures [latex]5[/latex] feet.Example
1. Use the multiplication and division properties of equality to isolate the variable [latex]b[/latex].Answer:
[latex]\begin{array}{l}\,\,\,\,\,\,\,\,A=\frac{1}{2}bh\\\\\left(2\right)A=\left(2\right)\frac{1}{2}bh\\\\\,\,\,\,\,\,2A=bh\\\\\,\,\,\,\,\,\,\frac{2A}{h}=\frac{bh}{h}\\\\\,\,\,\,\,\,\,\,\frac{2A}{h}=\frac{b\cancel{h}}{\cancel{h}}\end{array}[/latex]
Write the equation with the desired variable on the left-hand side as a matter of convention:[latex]b=\frac{2A}{h}[/latex]
2. Use the multiplication and division properties of equality to isolate the variable [latex]h[/latex].Answer:
[latex]\begin{array}{l}\,\,\,\,\,\,\,\,A=\frac{1}{2}bh\\\\\left(2\right)A=\left(2\right)\frac{1}{2}bh\\\\\,\,\,\,\,\,2A=bh\\\\\,\,\,\,\,\,\,\frac{2A}{b}=\frac{bh}{b}\\\\\,\,\,\,\,\,\,\,\frac{2A}{b}=\frac{h\cancel{b}}{\cancel{b}}\end{array}[/latex]
Write the equation with the desired variable on the left-hand side as a matter of convention:[latex]h=\frac{2A}{b}[/latex]
Think About It
Express the formula for the surface area of a cylinder, [latex]S=2\pi rh+2\pi r^{2}[/latex], in terms of the height, [latex]h[/latex]. In this example, the variable [latex]h[/latex] is buried pretty deep in the formula for the surface area of a cylinder. Using the order of operations, it can be isolated. Before you look at the solution, use the box below to write down what you think is the best first step to take, to isolate [latex]h[/latex]. [practice-area rows="1"][/practice-area]Answer: Isolate the term containing the variable, [latex]h[/latex], by subtracting [latex]2\pi r^{2}[/latex] from both sides.
[latex]\begin{array}{r}S\,\,=2\pi rh+2\pi r^{2} \\ \underline{-2\pi r^{2}\,\,\,\,\,\,\,\,\,\,\,\,\,-2\pi r^{2}}\\S-2\pi r^{2}\,\,\,\,=\,\,\,\,2\pi rh\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]
Next, isolate the variable [latex]h[/latex] by dividing both sides of the equation by [latex]2\pi r[/latex].[latex]\begin{array}{r}\frac{S-2\pi r^{2}}{2\pi r}=\frac{2\pi rh}{2\pi r} \\\\ \frac{S-2\pi r^{2}}{2\pi r}=h\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]
You can rewrite the equation so the isolated variable is on the left side.[latex]h=\frac{S-2\pi r^{2}}{2\pi r}[/latex]
Temperature
Let’s look at another formula that includes parentheses and fractions: the formula for converting from the Fahrenheit temperature scale to the Celsius scale.[latex]C=\left(F--32\right)\cdot \frac{5}{9}[/latex]
Example
Given a temperature of [latex]12^{\circ}{C}[/latex], find the equivalent in [latex]{}^{\circ}{F}[/latex].Answer: Substitute the given temperature in [latex]{}^{\circ}{C}[/latex] into the conversion formula:
[latex]12=\left(F-32\right)\cdot \frac{5}{9}[/latex]
Isolate the variable [latex]F[/latex] to obtain the equivalent temperature.[latex]\begin{array}{r}12=\left(F-32\right)\cdot \frac{5}{9}\\\\\left(\frac{9}{5}\right)12=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\\left(\frac{108}{5}\right)=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\21.6=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\underline{+32\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+32}\,\,\,\,\,\,\,\,\,\,\,\,\\\\53.6={}^{\circ}{F}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]
Example
Solve the formula shown below for converting from the Fahrenheit scale to the Celsius scale for [latex]F[/latex]. [latex-display]C=\left(F--32\right)\cdot \frac{5}{9}[/latex-display]Answer: To isolate the variable [latex]F[/latex], it would be best to clear the fraction involving [latex]F[/latex] first. Multiply both sides of the equation by [latex] \displaystyle \frac{9}{5}[/latex].
[latex]\begin{array}{l}\\\,\,\,\,\left(\frac{9}{5}\right)C=\left(F-32\right)\left(\frac{5}{9}\right)\left(\frac{9}{5}\right)\\\\\,\,\,\,\,\,\,\,\,\,\,\,\frac{9}{5}C=F-32\end{array}[/latex]
Add [latex]32[/latex] to both sides.[latex]\begin{array}{l}\frac{9}{5}\,C+32=F-32+32\\\\\frac{9}{5}\,C+32=F\end{array}[/latex]
Answer
[latex]F=\frac{9}{5}C+32[/latex]Try It
[ohm_question]2379[/ohm_question]Practice Isolating a Variable in a Formula
Try it
[ohm_question]196857[/ohm_question]Contribute!
Licenses & Attributions
CC licensed content, Original
- Find the Width of a Rectangle Given the Perimeter / Literal Equation. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Find the Base of a Triangle Given Area / Literal Equation. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Convert Celsius to Fahrenheit / Literal Equation. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Revision and Adaptation. Provided by: Lumen Learning License: CC BY: Attribution.
CC licensed content, Shared previously
- Unit 10: Solving Equations and Inequalities, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology Located at: https://www.nroc.org/. License: CC BY: Attribution.
