Chapter 11 · Area of polygons

Rectangles: Perimeter, Area, and Missing Sides

Find the perimeter and area of rectangles and squares, then work backward from an area or perimeter to find a missing side length.

16 min read

A rectangle is a four-sided figure with four right angles, which are square corners. Its opposite sides are equal in length. A square is a rectangle whose four sides are all equal. So every rule you learn here for rectangles also works for squares.

You can measure a rectangle in two different ways. One measure is the distance around its edge. The other is the amount of flat space inside it. In this lesson you will find both measures. Then you will work backward from a measure to a side length you do not know.

Start with the distance around. The perimeter of a figure is the total distance around it. Picture walking once around the edge and measuring how far you went. To find a perimeter, you add the lengths of all the sides. The lengths must be in the same unit, and the perimeter is in that same unit of length.

The figure below shows a rectangle on a coordinate plane. The horizontal axis is the xx-axis and the vertical axis is the yy-axis. Each point is named by a pair (x,y)(x, y): xx tells how far across the point is and yy tells how far up. One grid step stands for 11 centimeter. The corners of the rectangle, called vertices, are at (1,1)(1, 1), (7,1)(7, 1), (7,5)(7, 5) and (1,5)(1, 5).

The two bottom vertices share the yy-value 11, and the two top vertices share the yy-value 55. So the bottom and top sides are horizontal. The two left vertices share the xx-value 11, and the two right vertices share the xx-value 77. So the left and right sides are vertical. Horizontal and vertical grid lines are perpendicular, which means they meet at right angles. This figure has four right angles because of where its vertices are, not because of how it looks.

Coordinate grid from x 0 to 8 and y 0 to 6, with a grid step of 1, showing 1 plotted element. Details: polygon with 4 vertices.
The rectangle with vertices (1,1)(1, 1), (7,1)(7, 1), (7,5)(7, 5) and (1,5)(1, 5). One grid step is 11 centimeter.

Worked example

Perimeter of a rectangle on a grid

The rectangle in the figure has vertices (1,1)(1, 1), (7,1)(7, 1), (7,5)(7, 5) and (1,5)(1, 5). One grid step is 11 centimeter. What is its perimeter?

  1. Find the length of the bottom side, from (1, 1) to (7, 1). The side is horizontal, so its length is the difference of the x-values: subtract the smaller from the larger. One grid step is 1 centimeter, so the side is 6 centimeters long.

    7−1=67 - 1 = 6
  2. Find the length of the right side, from (7, 1) to (7, 5). The side is vertical, so its length is the difference of the y-values. The side is 4 centimeters long.

    5−1=45 - 1 = 4
  3. Opposite sides of a rectangle are equal, so the top side is also 6 centimeters and the left side is also 4 centimeters. Add all four sides, because the perimeter is the whole distance around.

    6+4+6+4=206 + 4 + 6 + 4 = 20
  4. Check with a shortcut. One 6-centimeter side and one 4-centimeter side take you halfway around. Double that sum, because the rectangle has two sides of each length.

    2×(6+4)=2×10=202 \times (6 + 4) = 2 \times 10 = 20

Answer

The perimeter of the rectangle is 2020 centimeters.

That shortcut works for every rectangle, because opposite sides of a rectangle are always equal. Let ll stand for the length of one side and ww for the length of a side next to it, both in the same unit. People often call ll the length and ww the width. Going once around, you pass ll, ww, ll and ww, so the perimeter PP is l+w+l+wl + w + l + w. You can group those four sides as (l+w)+(l+w)(l + w) + (l + w). That is two copies of l+wl + w, and regrouping does not change the total. So P=2×(l+w)P = 2 \times (l + w).

A square is a rectangle with ll and ww equal. Call its side length ss. Its perimeter is s+s+s+ss + s + s + s, which is 4×s4 \times s. For example, a square with sides of 55 centimeters has a perimeter of 4×5=204 \times 5 = 20 centimeters. The rectangle rule agrees: 2×(5+5)=202 \times (5 + 5) = 20.

The area of a figure is the amount of flat surface it covers. You measure area by counting unit squares. A unit square has sides 11 unit long, and the unit squares must cover the figure with no gaps and no overlaps. A square with sides of 11 centimeter has an area of 11 square centimeter, also written 1 cm21 \text{ cm}^2.

The figure below shows the same rectangle cut into four rows. The bottom row has vertices (1,1)(1, 1), (7,1)(7, 1), (7,2)(7, 2) and (1,2)(1, 2). Its sides are horizontal and vertical, so it is a rectangle too. It is 7−1=67 - 1 = 6 centimeters long and 2−1=12 - 1 = 1 centimeter tall. On this grid each cell is one unit square, so the row holds 66 unit squares. The other three rows lie between the yy-values 22 and 33, 33 and 44, and 44 and 55. Each of them also runs from the xx-value 11 to the xx-value 77, so each is 66 centimeters long and 11 centimeter tall.

Coordinate grid from x 0 to 8 and y 0 to 6, with a grid step of 1, showing 4 plotted elements. Details: polygon with 4 vertices; polygon with 4 vertices; polygon with 4 vertices; polygon with 4 vertices.
The same rectangle cut into 44 rows. Each row is 66 centimeters long and 11 centimeter tall, so it holds 66 unit squares.

The four rows cover the whole rectangle with no gaps and no overlaps. So the rectangle holds 4×6=244 \times 6 = 24 unit squares, and its area is 2424 square centimeters.

Notice where the 44 and the 66 come from. The rectangle is 66 centimeters long, so each row holds 66 unit squares. It is 44 centimeters wide, so there are 44 rows. The 44-centimeter side is perpendicular to the 66-centimeter side, so the rows stack straight up and fill the rectangle exactly. That is why you can find the area AA by multiplying the two side lengths instead of counting: A=6×4=24A = 6 \times 4 = 24 square centimeters.

A square works the same way, with both side lengths equal. A square with sides of 55 centimeters holds 55 rows of 55 unit squares, so its area is 5×5=255 \times 5 = 25 square centimeters. Earlier you found that its perimeter is 2020 centimeters. The two measures answer different questions, so they use different units.

Key concept

Perimeter goes around; area fills inside

Take a rectangle with length ll and width ww. These two sides meet at a right angle, and both are measured in the same unit.

  • The perimeter PP is the distance around the rectangle. Add two lengths and two widths. The answer is in units of length.
  • The area AA is the number of unit squares that cover the rectangle. Multiply the length by the width. The answer is in square units.

A square is a rectangle with l=w=sl = w = s, so P=4×sP = 4 \times s and A=s×sA = s \times s. These rules hold for whole-number, decimal and fraction side lengths. Each rectangle rule links three quantities. If you know two of them, you can find the third.

P=2×(l+w)A=l×wP = 2 \times (l + w) \qquad A = l \times w

Worked example

Both measures of one rectangle

A rectangle is 7.57.5 meters long and 44 meters wide. Find its perimeter and its area.

  1. Name what you know. Let l be the length and w be the width, both in meters. In a rectangle these two sides meet at a right angle, so both rules apply.

    l=7.5,w=4l = 7.5, \quad w = 4
  2. Find the perimeter. Add one length and one width to get halfway around. Then double the sum, because the rectangle has two sides of each length.

    P=2×(7.5+4)=2×11.5=23P = 2 \times (7.5 + 4) = 2 \times 11.5 = 23
  3. Find the area. Multiply the length by the width. Picture 4 rows that are each 1 meter tall. Each row holds 7 whole unit squares and half of another, so multiply the 7 and the 0.5 by 4 separately and add.

    A=7.5×4=(7×4)+(0.5×4)=28+2=30A = 7.5 \times 4 = (7 \times 4) + (0.5 \times 4) = 28 + 2 = 30
  4. Attach the units. The perimeter is a distance, so it is in meters. The area counts unit squares that are 1 meter on each side, so it is in square meters.

Answer

The rectangle has a perimeter of 2323 meters and an area of 3030 square meters.

Sometimes you know the area or the perimeter of a rectangle, but one side length is missing. You can still find it. Write the rule as an equation, which is a statement that two amounts are equal. Use a letter for the side length you do not know. Then undo the operations one at a time until the letter stands alone.

To undo an operation, use its inverse operation. Dividing by a nonzero number undoes multiplying by it, and subtracting a number undoes adding it. An equation stays true when you divide both sides of the equation by the same nonzero number, or subtract the same number from both sides. The two amounts change in the same way, so they stay equal.

Worked example

A missing width from the area

A rectangle has an area of 58.558.5 square meters. Its length is 99 meters. What is its width?

  1. Name the unknown. Let w be the width of the rectangle in meters. Write the area rule with the values you know: the length times the width equals the area.

    9×w=58.59 \times w = 58.5
  2. Undo the multiplication by 9. Divide both sides of the equation by 9. The two sides stay equal because you divide them by the same nonzero number.

    w=58.5÷9w = 58.5 \div 9
  3. Do the division in parts. Split 58.5 into 54 and 4.5, because each part divides evenly by 9. Then add the two results.

    58.5÷9=(54÷9)+(4.5÷9)=6+0.5=6.558.5 \div 9 = (54 \div 9) + (4.5 \div 9) = 6 + 0.5 = 6.5
  4. Check by multiplying the two side lengths. The result should be the area you started with.

    9×6.5=(9×6)+(9×0.5)=54+4.5=58.59 \times 6.5 = (9 \times 6) + (9 \times 0.5) = 54 + 4.5 = 58.5

Answer

The width of the rectangle is 6.56.5 meters.

Worked example

A missing width from the perimeter

A rectangle has a perimeter of 2626 centimeters. Its length is 88 centimeters. What is its width?

  1. Let w be the width of the rectangle in centimeters. Write the perimeter rule with the values you know: twice the sum of the length and the width equals the perimeter.

    2×(8+w)=262 \times (8 + w) = 26
  2. The rule adds first and doubles last, so undo the doubling first. Divide both sides of the equation by 2. What is left is one length plus one width, which is halfway around.

    8+w=26÷2=138 + w = 26 \div 2 = 13
  3. Undo the addition of 8. Subtract 8 from both sides of the equation, which keeps them equal.

    w=13−8=5w = 13 - 8 = 5
  4. Check by adding all four sides with the width you found.

    8+5+8+5=268 + 5 + 8 + 5 = 26

Answer

The width of the rectangle is 55 centimeters.

A square is quicker to work backward on, because its four sides are equal. Suppose a square has a perimeter of 3636 centimeters. Let ss be its side length in centimeters. Then 4×s=364 \times s = 36. Divide both sides of the equation by 44 to get s=36÷4=9s = 36 \div 4 = 9. Each side is 99 centimeters long. Check: 4×9=364 \times 9 = 36.

Here is the idea to keep. The perimeter of a rectangle is the distance around it, so you add: P=2×(l+w)P = 2 \times (l + w), in units of length. The area is the surface it covers, so you multiply two sides that meet at a right angle: A=l×wA = l \times w, in square units. Both rules need side lengths in the same unit, and a square follows both with equal sides. When a side length is missing, write the rule as an equation and undo each operation in turn.

Next, you will use these ideas to solve area and perimeter word problems and to compare the area and perimeter of two figures.