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# Area and Perimeter of Rectangles

## Area is base times height, while perimeter is the sum of the sides.

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Area and Perimeter of Rectangles
License: CC BY-NC 3.0

Kathryn is getting new carpeting for the three bedrooms in her house. She decided to get the same type of carpeting for each room. The carpeting she picked costs 3 per square foot. In order to place an order for the carpeting, she needs to know how much she will need. Luckily, each of her rooms are rectangles. The first bedroom is 15 feet by 12 feet. The second bedroom is 12 feet by 11 feet. The third bedroom is 9 feet by 10 feet. How many square feet of carpeting will Kathryn need to order? How much will it cost? In this concept, you will learn how to use formulas to find the perimeter and area of squares and rectangles. ### Finding Perimeter and Area of Rectangles Perimeter is the distance around a figure. A rectangle is a four-sided shape that has two lengths (l\begin{align*}l\end{align*}) that are equal and two widths (w\begin{align*}w\end{align*}) that are equal. The perimeter of a rectangle can be found using the formula below. Rectangle Perimeter:P=2l+2w\begin{align*}\text{Rectangle Perimeter}: P=2l+2w\end{align*} A square is a four-sided shape that has four equal sides (s\begin{align*}s\end{align*}). Therefore, it is possible to find the perimeter of a square by multiplying one of the side lengths by 4. The perimeter of a square can be found using the formula below. Square Perimeter:P=4s\begin{align*}\text{Square Perimeter}: P=4s\end{align*} Area is the number of square units contained inside a figure. The areas of rectangles and squares can be found using the formulas below. Rectangle Area:ASquare Area:A==lws2\begin{align*}\begin{array}{rcl} \text{Rectangle Area}: A &=& lw \\ \text{Square Area}: A &=& s^2 \end{array}\end{align*} Here is an example. Find the perimeter and area of the rectangle below. License: CC BY-NC 3.0 First, note the length and width of the rectangle. One of the sides is 12 inches and another is 7 inches. This will be your length and width. It does not actually matter which one you assign as length and which one you assign as width. l=12 in;w=7 in\begin{align*}l=12 \ in; w=7 \ in\end{align*} Next, find the perimeter by substituting the values for l\begin{align*}l\end{align*} and w\begin{align*}w\end{align*} into the perimeter formula and evaluating. PPPP====2l+2w2(12)+2(7)24+1438 in\begin{align*}\begin{array}{rcl} P & = & 2l+2w\\ P & = & 2(12)+2(7)\\ P & = & 24+14\\ P & = & 38 \ in \end{array}\end{align*} Notice that your units for perimeter are inches because the original dimensions of the rectangle were in inches. The units for perimeter will always be the same as the units of the dimensions. Now, find the area by substituting the values for l\begin{align*}l\end{align*} and w\begin{align*}w\end{align*} into the area formula and evaluating. AAA===lw(12)(7)84 in2 (or 84 square inches)\begin{align*}\begin{array}{rcl} A & = & lw\\ A & = & (12)(7)\\ A & = & 84 \ in^2 \ (\text{or }84 \text{ square inches}) \end{array}\end{align*} Notice that your units for area are square inches. The units for area will always be the square of the units of the dimensions. This is because area is a calculation of the number of square units it takes to cover a figure. The answer is the perimeter of the rectangle is 38 in and the area of the rectangle is 84 in2\begin{align*}84 \ in^2\end{align*}. Here is another example. Find the perimeter and area of the square below. License: CC BY-NC 3.0 First, note that only one dimension is given. This is because all four sides of a square are equal in length. s=14 ft\begin{align*}s=14 \ ft\end{align*} Next, find the perimeter by substituting the value for s\begin{align*}s\end{align*} into the perimeter formula and evaluating. PPP===4s4(14)56ft\begin{align*}\begin{array}{rcl} P & = & 4s\\ P & = & 4(14)\\ P & = & 56 ft \end{array}\end{align*} Now, find the area by substituting the value for s\begin{align*}s\end{align*} into the area formula and evaluating. AAA===s2142196 ft2 (or 196 square feet)\begin{align*}\begin{array}{rcl} A & = & s^2\\ A & = & 14^2\\ A & = & 196 \ ft^2 \ (\text{or } 196 \text{ square feet}) \end{array}\end{align*} The answer is the perimeter of the square is 56 in and the area of the square is 196 in2\begin{align*}196 \ in^2\end{align*}. ### Examples #### Example 1 Earlier, you were given a problem about Kathryn and her new carpets. Kathryn is ordering new carpeting for the three bedrooms in her house. The carpeting she picked costs3 per square foot. The bedrooms are 15 ft×12 ft,12 ft×11 ft\begin{align*}15 \ ft \times 12 \ ft, 12 \ ft \times 11 \ ft\end{align*}, and 9 ft×10 ft\begin{align*}9 \ ft \times 10 \ ft\end{align*}. Kathryn needs to know how many square feet of carpeting to order and how much it will cost.

First, find the area of each bedroom. Because the bedrooms are rectangles, you can use the formula A=lw\begin{align*}A=lw\end{align*} to find the area of each.

First Bedroom:

AAA===lw(15)(12)180 ft2\begin{align*}\begin{array}{rcl} A & = & lw\\ A & = & (15)(12)\\ A & = & 180 \ ft^2 \end{array}\end{align*}

Second Bedroom:

AAA===lw(12)(11)132 ft2\begin{align*}\begin{array}{rcl} A & = & lw\\ A & = & (12)(11)\\ A & = & 132 \ ft^2 \end{array}\end{align*}

Third Bedroom:

AAA===lw(9)(10)90 ft2\begin{align*}\begin{array}{rcl} A & = & lw\\ A & = & (9)(10)\\ A & = & 90 \ ft^2 \end{array}\end{align*}

Next, add up the three areas in order to find the total amount of carpet that Kathryn will need to order.

Total AreaTotal Area==180+132+90402 ft2\begin{align*}\begin{array}{rcl} \text{Total Area} & = & 180+132+90\\ \text{Total Area} & = & 402 \ ft^2 \end{array}\end{align*}

The first answer is that Kathryn will need to order 402 square feet of carpeting.

Now, find the total cost. The carpet costs $3 per square foot. Multiply the total amount of carpeting by$3 to find the total cost.

CostCost==402$3$1206\begin{align*}\begin{array}{rcl} \text{Cost} & = & 402 \cdot \3\\ \text{Cost} & = & \1206 \end{array}\end{align*}

The answer is it will cost \$1206 for Kathryn to get new carpeting for the bedrooms in her house.

#### Example 2

Marcy has purchased a new rug for her office. The rug measures 6 ft×9 ft\begin{align*}6 \ ft \times 9 \ ft\end{align*}. Given these measurements, what is the perimeter of the rug? What is the area?

First, notice that you were given the dimensions 6 ft×9 ft\begin{align*}6 \ ft \times 9 \ ft\end{align*} for the rug. From this you can deduce that the rug is a rectangle. The length is 6 feet and the width is 9 feet.

Next, find the perimeter of the rug using the perimeter of a rectangle formula.

PPPP====2l+2w2(6)+2(9)12+1830 ft\begin{align*}\begin{array}{rcl} P & = & 2l+2w\\ P & = & 2(6)+2(9)\\ P & = & 12+18\\ P & = & 30 \ ft \end{array}\end{align*}

Then, find the area of the rug using the area of a rectangle formula.

AAA===lw(6)(9)54 ft2\begin{align*}\begin{array}{rcl} A & = & lw\\ A & = & (6)(9)\\ A & = & 54 \ ft^2 \end{array}\end{align*}

The answer is that the perimeter of the rug is 30 feet and the area of the rug is 54 square feet.

#### Example 3

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

First, note that for this problem you have:

l=12 in;w=10 in\begin{align*}l=12 \ in; w=10 \ in\end{align*}

Now, find the perimeter by substituting the values for l\begin{align*}l\end{align*} and w\begin{align*}w\end{align*} into the perimeter formula and evaluating.

PPPP====2l+2w2(12)+2(10)24+2044 in\begin{align*}\begin{array}{rcl} P & = & 2l+2w\\ P & = & 2(12)+2(10)\\ P & = & 24+20\\ P & = & 44 \ in \end{array}\end{align*}

Next, find the area by substituting the values for l\begin{align*}l\end{align*} and w\begin{align*}w\end{align*} into the area formula and evaluating.

AAA===lw(12)(10)120 in2\begin{align*}\begin{array}{rcl} A & = & lw\\ A & = & (12)(10)\\ A & = & 120 \ in^2 \end{array}\end{align*}

The answer is the perimeter of the rectangle is 44 in and the area of the rectangle is 120 in2\begin{align*}120 \ in^2\end{align*}.

#### Example 4

Find the perimeter of a square with a side length of 4.5 inches.

First, note that only one dimension was given. This is because all four sides of a square are equal in length.

s=4.5 in\begin{align*}s=4.5 \ in\end{align*}

Next, find the perimeter by substituting the value for s\begin{align*}s\end{align*} into the perimeter formula and evaluating.

PPP===4s4(4.5)18 in\begin{align*}\begin{array}{rcl} P & = & 4s\\ P & = & 4(4.5)\\ P & = & 18 \ in \end{array}\end{align*}

The answer is the perimeter of the square is 18 in.

#### Example 5

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

First, note that for this problem you have:

l=15 ft;w=12 ft\begin{align*}l=15 \ ft;w=12 \ ft\end{align*}

Next, find the perimeter by substituting the values for l\begin{align*}l\end{align*} and w\begin{align*}w\end{align*} into the perimeter formula and evaluating.

PPPP====2l+2w2(15)+2(12)30+2454 ft\begin{align*}\begin{array}{rcl} P & = & 2l+2w\\ P & = & 2(15)+2(12)\\ P & = & 30+24\\ P & = & 54 \ ft \end{array}\end{align*}

The answer is the perimeter of the rectangle is 54 ft.

### Review

Find the area and perimeter of each rectangle 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

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

1. \begin{align*}s = 6 \ ft\end{align*}
2. \begin{align*}s = 8 \ ft\end{align*}
3. \begin{align*}s = 9 \ in\end{align*}
4.  \begin{align*}s= 4 \ in\end{align*}
5.  \begin{align*}s= 12 \ in\end{align*}
6.  \begin{align*}s= 7 \ ft\end{align*}
7.  \begin{align*}s= 5 \ cm\end{align*}
8.  \begin{align*}s= 3 \ m\end{align*}
9.  \begin{align*}s= 10 \ m\end{align*}
10.  \begin{align*}s= 11 \ yards\end{align*}

To see the Review answers, open this PDF file and look for section 1.14.

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### Vocabulary Language: English

Area

Area is the space within the perimeter of a two-dimensional figure.

Perimeter

Perimeter is the distance around a two-dimensional figure.

Area of a Rectangle

To find the area 'A' of a rectangle, calculate A = bh, where b is the base (width) and h is the height (length).

Perimeter of a Rectangle

The perimeter 'P' of a rectangle is equal to twice the base added to twice the height: P = 2b + 2h.