# 1.6: Using Formulas

Difficulty Level: At Grade Created by: CK-12

## Introduction

Shelter Repair at the Lonesome Lake Hut

On the morning after the groups were divided and the food was packed up, the groups headed into the backcountry. Kelly’s group began hiking up the Lonesome Lake Trail which starts at the base of the mountains in Lafayette Place Campground. They were carrying heavy gear, but were very excited.

Their group leaders, Scott and Laurel, told them that their first stop would be the Lonesome Lake Hut. The AMC has huts stationed all through the White Mountains. Some of them are even at the top of mountains. These huts have beds, some have full kitchens, some have self-service kitchens. Hikers of all different kinds stay at the huts. The Lonesome Lake Hut is located at 2,760 feet and the trail is 1.5 miles long. It would be one of the easier hikes they would go on.

Scott told the group that they would be helping to repair the floor of the main house. It seemed that there was some water damage. The roof had been repaired but the floor needed work.

The group hiked the beautiful, relatively easy trail in about an hour and a half. When they arrived, they were delighted to see everything. After unpacking in the bunkhouse and getting some lunch, the hikers went to work.

The crew leader at the hut told the students that the floor was damaged during the winter. The area of the floor was 90,720 square feet and one quarter of the floor needed to be fixed. That is an area of 22,680 square feet and a length of 90 ft.

If the length is 90 ft, and the area is 22,680 square feet, what is the width of the area that needs repair?

To figure this out, you will need to learn about formulas and working backwards. Pay attention and you will learn how to solve this problem by the end of the lesson.

What You Will Learn

By the end of this lesson you will be able to perform the following skills.

• Choose appropriate tools and units for given measurement situations involving squares and rectangles.
• Find perimeter and area of squares and rectangles using formulas.
• Solve for unknown dimensions using formulas, given perimeter or area.
• Solve real-world problems involving perimeter and area, including irregular figures made of rectangles and squares.

Teaching Time

I. Choose Appropriate Tools and Units for Given Measurement Situations Involving Squares and Rectangles

Measurement is the system of comparing an object to a standard. 1 pound, 1 yard, and 1 liter are just a few of the standards we use. Measuring the weight of a bunch of bananas against 1 pound, the length of a football field against 1 yard, or the amount of milk needed to make a cake against 1 liter helps us understand the quantities we encounter in all aspects of our lives.

We can measure the size, the weight or the capacity of different items.

When we measure, we can use two different measurement systems.

1. Customary System – the Customary System is one that you are very familiar with. We use it all the time here in the United States. The customary system uses units such as inch, foot, yard, pint, ounce, quart, pound and ton.
2. Metric System – the Metric System is a system used outside of the United States. Here in the US, we use often use metrics in science. Common units in the Metric System are millimeter, liter, gram and kilogram.

In this lesson, we will look at measuring length and distance, which are really the same thing but have different names based on how and where they are used.

Customary units of length/distance are inch, foot, yard and mile.

Metric units of length are millimeter, centimeter, meter and kilometer.

Measuring length always involves tools. There are a variety of tools to choose from when measuring length.

The goal in choosing a measurement tool and unit is to find the standard that most closely matches the thing you are measuring. For example, you wouldn’t want to measure the length of a city block in inches because the number would be too large and difficult to manage.

This table can give you an idea of how the different units of measurement relate to each other. Clearly an inch is smaller than a foot, just like a centimeter is smaller than a kilometer. Using this table will help you to choose appropriate units when measuring.

Customary Units of Length

inch (in)foot (ft)yard (yd)mile (mi)12 inches3 feet5,280 feet\begin{align*}& \text{inch} \ (in)\\ & \text{foot} \ (ft) && 12 \ inches\\ & \text{yard} \ (yd) && 3 \ feet\\ & \text{mile} \ (mi) && 5,280 \ feet\end{align*}

Metric Units of Length

\begin{align*}& \text{millimeter} \ (mm) && \frac{1}{10} \ centimeter\\ & \text{centimeter} \ (cm) && \frac{1}{100} \ meter\\ & \text{meter} \ (m)\\ & \text{kilometer} \ (km) && 1,000 \ meters\end{align*}

Example

Here are some items and the best units to use in measuring the length or width of them.

a) Width of a sandbox - it would be best measured in feet, yards or meters.

b) Length of a book - it would be best measured in inches or centimeters.

c) Width of a swimming pool - it would be best measured in feet, yards or meters.

d) The length of a city street - it would be best measured in miles or kilometers.

Once you have determined the appropriate unit of measurement, you have to move on to selecting an appropriate tool.

When measuring a small item, it is best to use a ruler whether it is in metrics or customary units. Rulers are used to measure inches, centimeters or millimeters.

What size measurements should be used? When we measure squares or rectangles (or any other shape), we will have to choose appropriate units of measure and appropriate tools. For example, look at this small square and determine which tool would be easiest to measure it with.

Using a ruler would make the best sense! We can then measure it in inches or centimeters.

If we had a large item, such as a pool, measuring the length and width would be quite different.

Looking at the entire pool, we can see that it is in the shape of a rectangle.

It would make sense to measure the perimeter or edge of the pool in feet or meters.

1M. Lesson Exercises

Name the appropriate unit to use in each measurement situation.

1. The distance of a road race
2. The length of a magazine
3. The width of a football field

Take a few minutes to check your work with a partner.

II. Find Perimeter and Area of Squares and Rectangles Using Formulas

Once we obtain measurements of the length and width of squares and rectangles, we are able to use formulas to find perimeter and area. Perimeter is the distance around a figure. Area is the measurement of the square units inside a figure.

Because perimeter is the distance around a figure, we can find the perimeter of a rectangle or square by adding the lengths of all four sides. A rectangle has two lengths that are equal and two widths that are equal, so the perimeter of a rectangle can be described by the formula \begin{align*}P = 2l + 2w\end{align*}. Remember that a square has four equal sides. Therefore, it is possible to find the perimeter of a square by multiplying one of the sides by 4. The formula for finding the perimeter of a square is \begin{align*}P = 4s\end{align*} where \begin{align*}s\end{align*} is the measurement of one side.

Area is the space inside a figure; it is measured in square units. To find the area of rectangle, we multiply the length times the width. \begin{align*}A = lw\end{align*}.

A square has four equal sides, so when finding the area of a square, all we need to do is multiply one of the sides by itself. \begin{align*}A = s^2\end{align*}.

Sure. Here they are once again.

\begin{align*}&\text{Rectangle} && P = 2l + 2w\\ &&& A = lw\\ &\text{Square} && P = 4s\\ &&& A = s^2\end{align*}

Write these formulas down in a notebook before continuing with the lesson.

Now let’s look at an example.

Example

Find the perimeter and area of the following rectangle.

First, let’s find the perimeter of the rectangle. To do this, we take our formula and substitute the length in for \begin{align*}l\end{align*} and the width in for \begin{align*}w\end{align*}.

\begin{align*}P &= 2l+2w\\ P &= 2(12)+ 2(7)\\ P &= 24+14\\ P &= 38 \ inches\end{align*}

Now, we can find the area of the rectangle. To do this, we take the formula for area and substitute the given values in for length and width.

\begin{align*}A &= lw\\ A &= (12)(7)\\ A &= 84 \ sq.in.\end{align*}

Use the formulas and find the perimeter and area of the following square.

Here you can see that we have only been given the length on one side. That is okay though because this is a square. A square has four congruent or equal sides. Therefore, all we need is one side length to work with.

First, let’s find the perimeter of the square.

\begin{align*}P &= 4s\\ P &= 4(14)\\ P &= 56 \ feet\end{align*}

Next, we can use the formula for area to find the area of the square.

\begin{align*}P &= s^2\\ P &= 14^2=14(14)\\ P &= 196 \ square \ feet \ or \ ft^2\end{align*}

1N. Lesson Exercises

Find the perimeter and area of a rectangle with a width of 10 inches and length of 12 inches.

Take a few minutes to check your work with a friend.

III. Solve for Unknown Dimensions Using Formulas, Given Perimeter and Area

You just learned how to find the perimeter and area given side measurements, now you are going to work backwards.

Remember, an equation states that two expressions are equal and a variable equation is an equation that includes an algebraic unknown, or a variable. Our formulas for perimeter and area are equations. In the examples above, the perimeter \begin{align*}(P)\end{align*} and the area \begin{align*}(A)\end{align*} were the variables. In some cases, if we already know the perimeter or the area, we can make the length or width our variable and solve for that value.

Example

The width of a rectangle is 10 feet and the perimeter is 50 feet. What is the length of the rectangle?

Here we are trying to find the length of the rectangle, so we’ll give that value the variable \begin{align*}l\end{align*}. We already know the perimeter and the width. All we have to do is substitute the values for the perimeter and the width into our formula and solve for \begin{align*}l\end{align*}.

Remember to follow the order of operations: Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.

\begin{align*}P &= 2l + 2w\\ 50 &= 2l + 2(10)\\ 50 &= 2l + 20\end{align*}

Now we can use mental math, we need a number plus 20 that will give us 50. 30 is a likely choice. Now we have \begin{align*}2l\end{align*} so that means 2 times some number. We want this product to be 30, so \begin{align*}l=15\end{align*}.

Let’s check our work by substituting 15 in for length. If both sides of the equation are equal, then our equation balances and our work is done.

\begin{align*}P &= 2l+2w\\ 50 &= 2(15)+ 2(10)\\ 50 &= 30+20\\ 50 &= 50\end{align*}

Our equation balances and our work is done.

What about finding the side length when you have the area of a square?

Let’s look at an example.

Example

If the area of a square is 144 sq. miles, what is the length of one of the sides of the square?

To do this, we work backwards and substitute the given area into the formula.

\begin{align*}A &= s^2\\ 144 &= s^2\end{align*}

Now we use mental math and ask ourselves the question "What number times itself is equal to 144?" The answer is 12.

The side length is 12 miles.

We can check our work by substituting the given values into the formula.

\begin{align*}144 &= 12^2\\ 144 &= 144\end{align*}

Our equation balances and our work is accurate.

IV. Solve Real-World Problems Involving Perimeter and Area, Including Irregular Figures Made of Rectangles and Squares

Now you can apply all of these skills to real-world problems. Let’s look at an example.

Example

The perimeter of a square playground is 200 yards. What is the length of one of the sides of the playground?

This is one of those "working backwards" problems. Fill the given information into the formula and solve for the missing side length.

\begin{align*}P &= 4s\\ 200 &= 4s\end{align*}

What number times four is equal to 200? 50 is the correct answer. \begin{align*}50 \times 4 = 200\end{align*}

The side length of one side of the playground is 50 yards.

We can also use this information to find the area of a figure that might be made up of more than one square or rectangle. These figures are called “irregular shapes.”

Look at the following diagram.

Example

The Hegazzi family is designing a summer garden based on the model shown above. Plot \begin{align*}A\end{align*} is a square and plot \begin{align*}B\end{align*} is a rectangle. If the total area of both plots in the garden is 139 square feet, what is the length of one side of plot \begin{align*}A\end{align*}?

To solve this multi-step problem, we will have to find the area of plot \begin{align*}B\end{align*} and subtract it from the total area to find the area of plot \begin{align*}A\end{align*}. Then, since we know plot \begin{align*}A\end{align*} is a square, we will have to determine what length multiplied by itself will give us the length of one side of plot \begin{align*}A\end{align*}.

Let’s get started. Plot \begin{align*}B\end{align*} is a rectangle. The formula for the area of a rectangle is \begin{align*}A = lw\end{align*}.

\begin{align*}A &= lw\\ A &= 15(6)\\ A &= 90 \ ft^2\end{align*}

So the area of plot \begin{align*}B\end{align*} is \begin{align*}90 \ ft^2\end{align*}. We know the total area of the garden is \begin{align*}139 \ ft^2\end{align*}, but we need to find the area of plot \begin{align*}A\end{align*}.

We can use the following equation: Total area = area of plot \begin{align*}A\end{align*} + area of plot \begin{align*}B\end{align*}. Let’s assign the value of the area of plot \begin{align*}A\end{align*} the variable \begin{align*}x\end{align*}.

Total area = area of plot \begin{align*}A\end{align*} + area of plot \begin{align*}B\end{align*}.

\begin{align*}139 = x + 90\end{align*}

“What number plus 90 is 139?” We can figure this out by subtracting 90 from 139. The answer is 49.

\begin{align*}x=49\end{align*}

Now we know the area of plot \begin{align*}A = 49 \ ft^2\end{align*}, but we are looking for the length of one side. Plot \begin{align*}A\end{align*} is a square and all the sides are equal. The formula for the area of a square is \begin{align*}A = s^2\end{align*}. Let’s see how it looks:

\begin{align*}A &= s^2\\ 49 &= s^2\end{align*}

We need to think of what number times itself is equal to 49. \begin{align*}7 \times 7 = 49\end{align*}, so the length of one side of plot \begin{align*}A\end{align*} is 7 ft.

## Real Life Example Completed

Shelter Repair at the Lonesome Lake Hut

Here is the original problem once again. Reread the problem and then underline any important information.

On the morning after the groups were divided and the food was packed up, the groups headed into the backcountry. Kelly’s group began hiking up the Lonesome Lake Trail which starts at the base of the mountains in Lafayette Place Campground. They were carrying heavy gear, but were very excited. Their group leaders, Scott and Laurel, told them that their first stop would be the Lonesome Lake Hut. The AMC has huts stationed all through the White Mountains. Some of them are even at the top of mountains. These huts have beds, some have full kitchens, some have self-service kitchens. Hikers of all different kinds stay at the huts. The Lonesome Lake Hut is located at 2,760 feet and the trail is 1.5 miles long. It would be one of the easier hikes they would go on. Scott told the group that they would be helping to repair the floor of the main house. It seemed that there was some water damage. The roof had been repaired but the floor needed work. The group hiked the beautiful and relatively easy trail in about an hour and a half. When they arrived, they were delighted to see everything. After unpacking in the bunkhouse and getting some lunch, the hikers went to work.

The crew leader at the hut told the students that the floor was damaged during the winter. The area of the floor was 90,720 square feet and one quarter of the floor needed to be fixed. That is an area of 22,680 square feet with a length of 90 ft.

If the length is 90 ft and the area is 22,680 square feet, what is the width of the area that needs repair?

First, let’s write an equation to solve this problem. We can assume that the hut is a rectangle in shape because we were given a length and asked to find the width.

\begin{align*}A &= lw\\ 22680 &= 90w\end{align*}

We need to find a number that, when multiplied by 90, is equal to 22,680. To do this, it makes sense to divide. Division is the opposite of multiplication. Since mental math won’t work for this one, division is the next best option.

\begin{align*}22680 \div 90 = 63 \ ft\end{align*}

The width of the area of damaged floor is 63 feet wide.

The students worked hard to repair the floor and it was a great two days working at Lonesome Lake Hut! Yet when their work was done they were excited to think about hiking to their next destination!!

## Vocabulary

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

Measurement
the system of comparing an object to a standard.
Customary System
the system of measurements that uses inches, feet and miles, to name a few.
Metric System
the system of measurement that uses centimeters, meters and kilometers, to name a few.
Perimeter
the distance around the edge of a figure
Area
the measurement of the space inside a figure
Equation
one expression equals another expression
Variable Equation
an equation where a variable is used to represent an unknown quantity.

Resources

Resources

## Time to Practice

Directions: Find the area and perimeter of each rectangle by using the given dimensions.

1. Length = 10 in, width = 5 in

2. Length = 12 ft, width = 8 feet

3. Length = 11 ft, width = 5 feet

4. Length = 17 miles, width = 18 miles

5. Length = 22 ft, width = 20 feet

6. Length = 8 cm, width = 6 cm

7. Length = 20 cm, width = 17 cm

8. Length = 3 feet, width = 2 feet

9. Length = 15 yards, width = 11 yards

10. Length = 10 yards, width = 6 yards

Directions: Find the area and perimeter of each square using the given dimensions.

11. \begin{align*}s = 6 \ ft\end{align*}

12. \begin{align*}s = 8 \ ft\end{align*}

13. \begin{align*}s = 9 \ in\end{align*}

14. \begin{align*}s = 4 \ in\end{align*}

15. \begin{align*}s = 12 \ in\end{align*}

16. \begin{align*}s = 7 \ ft\end{align*}

17. \begin{align*}s = 5 \ cm\end{align*}

18. \begin{align*}s = 3 \ m\end{align*}

19. \begin{align*}s = 10 \ m\end{align*}

20. \begin{align*}s = 11 \ yards\end{align*}

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