<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# 1.6: Using Formulas

Difficulty Level: At Grade Created by: CK-12

## Introduction

Summer Travels

“What did you do this summer for fun?” Kevin asked Laila at lunch during the first week of school.

“We went camping at Yellowstone National Park and it was the best,” Laila said taking a bite of her sandwich.

“Really, that must have taken forever,” Kevin commented.

“No, we flew into Denver and then drove from Denver to Yellowstone so that we could see some of the country.”

“How far is it from Denver to Yellowstone?”

“It is about 540 miles. I know because I tracked it on the map. After leaving Denver it only took us nine hours to get there. We just made little stops and drove straight through so we could get there in time,” Laila explained.

Kevin began figuring out the math in his head. He wondered how fast the car was traveling if it made it in just nine hours.

Figuring out this problem will require you to use formulas. Formulas are often used in math as a method for solving a particular type of problem. Pay close attention to this lesson and you will see this problem once again.

What You Will Learn

By the end of this lesson, you will have an understanding of how to use the following skills.

• Find perimeter and area of squares and rectangles using formulas.
• Solve for unknown dimensions using formulas, given perimeter or area.
• Recognize and rewrite the formula for distance as the product of rate and time to find distance, to find rate, and to find time.
• Solve real-world problems involving simple familiar formulas, including area, distance and unit conversions.

Teaching Time

I. Find Perimeter and Area of Squares and Rectangles using Formulas

This lesson is all about formulas. Let’s start by thinking about what a formula actually is. A formula is a method that has been proven to work when solving specific types of problems. By this point in mathematics, there are many formulas that you are already familiar with. This lesson will explore some of those familiar formulas and their applications in problem solving. Let’s start by looking at rectangles, squares, area and perimeter.

The perimeter of a figure is the distance around the figure. Perimeter is the sum of all of the sides in a square or rectangle. Since a rectangle has two sets of parallel sides, the formula for determining perimeter of a rectangle is: \begin{align*}2L + 2W\end{align*}. \begin{align*}L =\end{align*} length and \begin{align*}W =\end{align*} width.

We can solve for the perimeter of this rectangle by substituting the given values for the variables that represent length and width.

\begin{align*}2(9) + 2(12) &= \text{Perimeter}\\ 18 + 24 &= \text{Perimeter}\\ 42 \ inches &= \text{Perimeter}\end{align*}

The perimeter of this rectangle is 42 inches.

Area is the amount of square units inside the figure. Area is found by multiplying the \begin{align*}\text{Length} \times \text{Width}\end{align*}. The formula for finding the area of a rectangle is \begin{align*}L \times W\end{align*}.

We can use the dimensions from the rectangle above to find the area of this rectangle.

\begin{align*}12 \times 9 &= \text{Area}\\ 12 \times 9 &= \text{Area}\\ 108 \ cm^2 &= \text{Area}\end{align*}

The area of this rectangle is \begin{align*}108 \ cm^2\end{align*}.

Copy these two formulas and the note about square units in your notebook. Use a drawing if necessary to help you remember each formula.

We can also find the perimeter and area of a square. Remember that a square has four equal sides, so we can use the following formula for finding the perimeter of a square, \begin{align*}4s\end{align*}.

Example

Find the perimeter of the following square.

Now we use the measurement of the side and the formula for finding the perimeter of a square.

\begin{align*}4s &= \text{Perimeter}\\ 4(4.5) &= \text{Perimeter}\\ 18 \ inches &= \text{Perimeter}\end{align*}

The perimeter of the square is 18 inches.

We can also find the area of a square. To find the area of a square we simply multiply the measurement of the side by itself.

\begin{align*}A= s^2\end{align*}

We know the length of the side, so we simply substitute it into the formula and solve.

\begin{align*}A &= (4.5)(4.5)\\ A &= 20.25 \ in^2\end{align*}

The area of the square is \begin{align*}20.25 \ in^2\end{align*}.

Take a few minutes to write the formulas for perimeter and area of a square down in your notebooks.

II. Solve for Unknown Dimensions using Formulas, Given Perimeter or Area

In the last section, you solved for the perimeter or area of a square or rectangle using the given side lengths. In this section, you are going to be working backwards by using the perimeter or area to figure out a missing length of a square or rectangle.

Example

A rectangle has a length of 12 feet and a perimeter of 72 feet. Write and solve an equation to determine the width of the rectangle.

Recall that the formula to determine perimeter of a rectangle is \begin{align*}2(L) + 2(W)\end{align*}. Substitute the information given in the problem to determine the missing dimension. We know the length of one side is 12 feet. Therefore, plug in 12 for the length. We also know that the total perimeter is 72 feet. Therefore, set the formula equal to 72 feet. Then, use inverse operations to solve for the unknown width.

\begin{align*}\text{Perimeter} &= 2(L) + 2(W)\\ \text{Perimeter} &= 2(12)+ 2(W)= 72\\ \text{Perimeter} &= 24 + 2W = 72\end{align*}

Next, we need to work to figure out the value of the width. It makes sense to use an inverse operation and subtract 24 from 72.

\begin{align*}2W=48\end{align*}

Two times twenty-four is forty-eight. This is our answer.

The unknown width is 24 feet.

Example

A square has a perimeter of 196 inches. Determine the length of one side of the square.

Recall that a square has four equal sides. Therefore, \begin{align*}4s = 196 \ inches\end{align*}. Use inverse operations to solve for “\begin{align*}s\end{align*}.”

\begin{align*}\frac{4s}{4} &= 196\\ s &= 49 \ inches\end{align*}

You can check your work by substituting 49 for the variable in \begin{align*}4s\end{align*}. 49 is the correct length if your answer is 196 inches.

\begin{align*}4s &= 196\\ 4(49) &= 196\\ 196 &= 196\end{align*}

The length of one side of the square is 49 inches.

Example

The area of a rectangle is 240 square feet. The length of one side is 15 feet. Write and solve an equation to determine the width of one side of the rectangle.

Recall that the formula for area is \begin{align*}L \times W\end{align*}. Plug in the information given in the problem. Then, use algebra to solve for the unknown width.

\begin{align*}L \times W &= Area\\ 15W &= 240\end{align*}

To figure this out, we divide 240 by 15. This is an example of using an inverse operation.

\begin{align*}W = 16 \ feet\end{align*}

We can check our work by substituting 16 for the width in the equation \begin{align*}L \times W = Area\end{align*}.

\begin{align*}15 \times 16 &= \text{Area}\\ 15 \times 16 &= 240 \ square \ feet\end{align*}

The rectangle is 15 feet by 16 feet.

You can see that by using the same formulas for finding the area and perimeter of squares and rectangles that we can work backwards and figure out missing parts. Remember it is a lot like figuring out the pieces of a puzzle.

III. Recognize and Rewrite the Formula for Distance as the Product of Rate and Time to Find Distance, to Find Rate and to Find Time

In the last section, you looked at measuring the distance around a figure and the distance inside a figure. We also measure distance-as in how far you can travel by car, boat, train or foot. We can measure rate or the speed that you can travel by each different mode, and finally we can measure the time or how long that it takes us. Look at the following example and then let’s look at how we can calculate distance, rate and time.

Example

The Murphy Family drove for three and a half hours from Manhattan, New York to Providence, Rhode Island at a rate of fifty-three miles per hour. Determine the distance at which the Murphy’s were traveling.

First, we need to think about what we need to solve for. In this problem, we need to figure out the distance that the Murphy family traveled. To do this, there is a simple formula that we can use. In fact, we can use this formula whenever we are calculating distance.

\begin{align*}Distance = rate \times time\end{align*}

Now we can take the given information from the problem and substitute that information into the formula. Once we have done that, we will be able to calculate the distance.

\begin{align*}D=53(3.5)\end{align*}

53 represents the rate or speed that the car was traveling.

3.5 represents the three and a half hours that the family traveled.

\begin{align*}D=185.5 \ miles\end{align*}

The distance that the Murphy family traveled was 185.5 miles.

That is a great question and the answer is “sort of”. You can use the same pieces of the formula only you will need to rewrite it to help you with the math. Here are two different versions of the same formula that you can use when looking to find the rate or the time.

To find rate, divide both sides of the equation by time.

\begin{align*}\text{Rate} = \frac{Distance}{Time}\end{align*}

To find time, divide both sides of the equation by rate.

\begin{align*}\text{Time} = \frac{Distance}{Rate}\end{align*}

Write these three formulas down in your notebooks.

Now let’s look at applying these formulas when problem solving.

Example

Use the distance formula to solve for rate.

\begin{align*}\text{Distance} = 285 \ miles\end{align*}

\begin{align*}\text{Time} = 9.5 \ hours\end{align*}

\begin{align*}\text{Rate} = x\end{align*}

Step 1: Recall that \begin{align*}\text{Rate} = \frac{\text{Distance}}{\text{Time}}\end{align*}.

\begin{align*}\text{Rate} = \frac{Distance}{Time}\end{align*}

Step 2: Substitute in the information given in the problem.

\begin{align*}\text{Rate} = \frac{285}{9.5}\end{align*}

Step 3: Solve

\begin{align*}\text{Rate} = 30 \ mph\end{align*}

Example

Use the distance formula to solve for time.

\begin{align*}\text{Distance} = 550 \ miles\end{align*}

\begin{align*}\text{Rate} = 55 \ mph\end{align*}

\begin{align*}\text{Time} = x\end{align*}

Step 1: Recall that \begin{align*}\text{Time} = \frac{\text{Distance}}{\text{Rate}}\end{align*}.

\begin{align*}\text{Time} = \frac{Distance}{Rate}\end{align*}

Step 2: Substitute in the given information into the equation.

\begin{align*}\text{Time} = \frac{550}{55}\end{align*}

Step 3: Solve

\begin{align*}\text{Time} = 10 \ hours\end{align*}

Now let’s look at how we can use all of these formulas when solving real-world problems.

IV. Solve Real – World Problems Involving Simple Familiar Formulas including Area, Distance and Unit Conversions

Formulas are used all the time in everyday life. If you think about what the formula helps you solve, you will be able to apply them in situations that make sense. Let’s work through a few examples and use formulas to solve real-world problems.

Example

The students at Horizon Elementary School are constructing a \begin{align*}400ft^2\end{align*} vegetable garden. The width of the garden is 25 feet. Determine the length of the vegetable garden.

Recall that \begin{align*}\text{Area} = \text{Length} \times \text{Width}\end{align*}. To determine the unknown length, plug the known information into the equation.

\begin{align*}Area &= Length \times Width\\ \frac{400}{25} &= \frac{25}{25} W\end{align*}

The width of the garden is 16 feet.

Example

A train traveled 255 miles in 300 minutes. Determine the rate at which the train was traveling.

Step 1: When using the distance formula, it is essential that you pay attention to the units used in the problem. If the rate in the problem gives miles per hour (mph), then time must be in hours. If the time is given in minutes, divide by sixty to determine the number of hours prior to solving the equation. Therefore prior to solving the problem above, divide 300 minutes by 60.

\begin{align*}300 \ minutes \div 60 \ minutes = 5 \ hours\end{align*}

Step 2: Since the problem is asking you to determine the rate, use the formula \begin{align*}\text{Rate} = \frac{\text{Distance}}{\text{Time}}\end{align*}. Plug the known information into the problem.

\begin{align*}\text{Rate} = \frac{Distance}{Time}\end{align*}

Step 3: Solve

\begin{align*}\text{Rate} &= \frac{255}{5}\\ \text{Rate} &= 51 \ miles \ per \ hour\end{align*}

The train was traveling at a rate of 51 miles per hour.

Example

A snake slithers at a rate of 2.5 feet per second. Determine the amount of time (in minutes) it will take a snake to slither a distance one mile.

Recall that \begin{align*}\text{Time} = \frac{\text{Distance}}{\text{Rate}}\end{align*}. Prior to solving the above problem, convert miles to feet. There are 5,280 feet in one mile. Therefore, use 5,280 feet to represent the distance traveled in the equation.

\begin{align*}\text{Time} &= \frac{Distance}{Rate}\\ \text{Time} &= \frac{5280 \ feet}{2.5 \ feet}\\ \text{Time} &= 2,112 \ seconds\end{align*}

The problem asks you to determine the amount of time in minutes. Therefore, convert seconds to minutes. There are 60 seconds in one minute. To convert the time to hours, divide 2,112 by 60.

\begin{align*}2,112 \ seconds \div 60 \ seconds = 35.2 \ minutes\end{align*}

Rounded to the nearest minute, the snake traveled thirty-five minutes.

## Real-Life Example Completed

Summer Travels

Here is the original problem once again. Reread it and then write the formula that you will need to solve it. After that, solve the problem for the speed of travel. Your answer will have two parts to it.

“What did you do this summer for fun?” Kevin asked Laila at lunch during the first week of school.

“We went camping at Yellowstone National Park and it was the best,” Laila said taking a bite of her sandwich.

“Really, that must have taken forever,” Kevin commented.

“No, we flew into Denver and then drove from Denver to Yellowstone so that we could see some of the country.”

“How far is it from Denver to Yellowstone?”

“It is about 540 miles. I know because I tracked it on the map. After leaving Denver it only took us nine hours to get there. We just made little stops and drove straight through so we could get there in time,” Laila explained.

Kevin began figuring out the math in his head. He wondered how fast the car was traveling if it made it in just nine hours.

Now write a formula and solve the problem. Remember, there are two parts to your answer.

Solution to Real – Life Example

To solve this problem you will need to use the formula:

\begin{align*}D=R \times T\end{align*}

Now that you have identified the correct formula, let’s think about the information that you have been given.

You know the distance from Denver to Yellowstone. It is 540 miles.

You know that it took the family 9 hours to get there with little quick stops.

We are looking to find the speed the car traveled.

Let’s substitute the given information into the formula.

\begin{align*}540=R(9)\end{align*}

Next, we can solve this problem by solving the equation. We want to figure out “\begin{align*}R\end{align*}” so we divide both sides of the equation by nine.

\begin{align*}\frac{540}{9} &= R\\ 60 \ mph &= R\end{align*}

The family was traveling about 60 mph. We could say that they were even traveling a little bit slower because there were quick stops.

## Vocabulary

Here are the vocabulary words that are found in this lesson.

Formula
a method proven to work when solving specific types of problems.
Perimeter
the distance around a figure.
Area
the measurement of the inside of a figure.
Distance
how far a vehicle or person travels in a certain amount of time given a specific rate.
Rate
the speed of travel
Time
the time of travel

## Time to Practice

Directions: Find the area and perimeter of each square or rectangle using formulas and the given dimensions. Each problem will have two answers.

1. A square with a side length of 5 inches.
2. A rectangle with a length of 5 inches and a width of 3 inches.
3. A rectangle with a length of 8 cm and a width of 6 cm.
4. A square with a side length of 11 feet.
5. A rectangle with a length of 9 inches and a width of 4.5 inches.
6. A square with a side length of 7 feet.
7. A rectangle with a length of 12 meters and a width of 11 meters.
8. A square with a side length of 13 meters.
9. A rectangle with a length of 15 feet and a width of 8 feet.
10. A square with a side length of 12.5 feet.

Directions: Find the missing side length given the area of each square.

1. \begin{align*}A = 64 \ in^2\end{align*}
2. \begin{align*}A = 36 \ in^2\end{align*}
3. \begin{align*}A = 81 \ m^2\end{align*}
4. \begin{align*}A = 100 \ in^2\end{align*}
5. \begin{align*}A = 144 \ ft^2\end{align*}
6. \begin{align*}A = 121 \ cm^2\end{align*}
7. \begin{align*}A = 4 \ mm^2\end{align*}

Directions: Find the number of miles traveled given the rate and time.

1. Four hours at a rate of 33 mph.
2. Six hours at a rate of 55 mph.
3. Eight hours at a rate of 65 mph.
4. 12 hours at a rate of 50 mph.
5. 14 hours at a rate of 60 mph.
6. 9 hours at a rate of 70 mph.

Directions: Use the given information to figure out each rate or time.

1. A car traveled 450 miles at a speed of 30 mph. How many hours did it take?
2. A car traveled 600 miles in 12 hours. What was the speed of the car?
3. A runner traveled 6 miles in 30 minutes. How fast was the runner going?
4. A car traveled 520 miles at a speed of 65 mph. How many hours did it take?

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes

Show Hide Details
Description
Difficulty Level:
Tags:
Subjects: