# 1.14: Area and Perimeter of Rectangles

**At Grade**Created by: CK-12

**Practice**Area and Perimeter of Rectangles

On the morning after the groups were decided and the food was packed up, the groups split up 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 tons of 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, but 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. 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 seems that there was some water damage. The roof has been repaired but the floor needs work.

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

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

If the length is 90 ft wide 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 know how to solve this problem at the end of the Concept.**

### Guidance

Measurement can also be completed with specific figures and 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.

What is perimeter and area? What does each of them measure?

** Perimeter** is the distance around a figure.

** Area** is 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*}

** Area** is the square units

*inside*a figure; it describes how much space a rectangle or square takes up. 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 Concept.*

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 measure for length in for \begin{align*}l\end{align*} and the measure for 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*}

Now try a few on your own.

#### Example A

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

**Solution: 44 inches**

#### Example B

The perimeter of a square with a side length of 4.5 inches.

**Solution: 18 inches**

#### Example C

The perimeter of a rectangle with a length of 15 feet and a width of 12 feet.

**Solution: 54 feet**

Now back to the shelter at Lonesome Lake. Here is the original problem once again.

On the morning after the groups were decided and the food was packed up, the groups split up 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 tons of 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, but 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. 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 seems that there was some water damage. The roof has been repaired but the floor needs work.

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

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

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

First, we 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 likely 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 in this Concept.

- Perimeter
- the distance around the edge of a figure

- Area
- the measurement inside a figure

### Guided Practice

Here is one for you to try on your own.

Marcy has purchased a new rug for her bedroom. The rug measures 6 ft. x 9 ft. Given these measurements, what is the perimeter of the rug? What is the area?

**Answer**

Because you were given the measurements 6' x 9', you can deduce that this rug is rectangle. Therefore we can find the perimeter of the rug by doubling the length and width and then add them together.

\begin{align*}P = 2(6) + 2(9)\end{align*}

\begin{align*}P = 30\end{align*} feet

The area of the rug can be found by using the following formula.

\begin{align*}A = lw\end{align*}

\begin{align*}A = (6)(9)\end{align*}

\begin{align*}A = 54\end{align*} sq. feet.

### Video Review

Here are videos for review.

- This is a James Sousa video on how to find the perimeter of a given rectangle.

- This is a James Sousa video on how to determine the area of a given rectangle.

### 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*}

### Image Attributions

Here you'll learn to use formulas to find the perimeter and area of squares and rectangles.