Choose Area or Perimeter and Compare Rectangles | Grade 6 Math | LevelKite

A rectangle is a closed figure with four straight sides and four right angles. A square is a rectangle too: it is a rectangle whose four sides are all equal. So everything in this lesson also works for squares.

You can measure a rectangle in two different ways.

Perimeter is the distance all the way around the edge. You measure it in length units, such as meters.
Area is the amount of flat surface inside the edge. You measure it in square units, such as square meters. One square meter is the surface of a square that is one meter long on each side.

A word problem usually does not tell you which measure to find. You have to decide. In this lesson you will pick the right measure for a story. Then you will compare both measures for two rectangles.

Key concept

Edge or surface? Choose the measure, then calculate

Let $\ell$ be the length and $w$ be the width of a rectangle, in the same length unit. These sides meet at a right angle, so they are perpendicular: $\ell$ is the base and $w$ is the height at a right angle to it.

Perimeter $P$ is the distance around the whole edge. Opposite sides are equal, so you pass $\ell$ , $w$ , $\ell$ , $w$ : two copies of $\ell + w$ . So $P = 2(\ell + w)$ length units.
Area $A$ is the surface inside, counted in squares $1$ unit on each side: $A = \ell \times w$ square units.

Both rules hold for every rectangle, squares included. In a story, ask: does this run along the whole edge, or cover the whole surface? The two measures are different things, so to compare two rectangles, find both for each one.

P = 2(\ell + w) \qquad A = \ell \times w

Ask what the story measures

What the story describes	Measure	Rule for a rectangle	Units of the answer
Something that runs along the whole edge, such as a fence, a picture frame or a ribbon border	Perimeter	$P = 2(\ell + w)$	Length units, such as meters (m)
Something that covers the whole surface, such as paint, carpet, tiles or grass	Area	$A = \ell \times w$	Square units, such as square meters ( $\text{m}^2$ )

Worked example

One garden, two questions

A rectangular garden is $8$ meters (m) long and $5$ m wide. You want to build a fence all the way around it and spread wood chips over all of the soil. How many meters of fence do you need? How many square meters ( $\text{m}^2$ ) of soil will the wood chips cover?

Name the two sides with their unit. The length is 8 m and the width is 5 m. The garden is a rectangle, so these sides meet at a right angle.
$\ell = 8 \text{ m}, \quad w = 5 \text{ m}$
Decide what the fence measures. The fence runs along the whole edge, so its length is the distance around the garden. That distance is the perimeter, measured in meters.
Find the perimeter. Going around, you pass the length, the width, then the length and the width again. That is the sum of one length and one width, taken twice. So add them and double the sum.
$P = 8 + 5 + 8 + 5 = 2(8 + 5) = 2 \times 13 = 26 \text{ m}$
Decide what the wood chips measure. They cover the whole surface inside the edge. That surface is the area, measured in square meters.
Find the area. Multiply the length by the width. This counts the one-meter squares that cover the soil: 5 rows with 8 squares in each row.
$A = 8 \times 5 = 40 \text{ m}^2$

Answer

You need $26$ meters of fence, and the wood chips will cover $40$ square meters of soil.

Worked example

Find the missing side, then the area

The boundary line painted all the way around a rectangular basketball court is $86$ meters (m) long. The court is $28$ m long. A crew will lay new flooring over the whole court. How many square meters ( $\text{m}^2$ ) of flooring does the crew need?

Sort the facts. The boundary line goes around the whole edge, so 86 m is the perimeter. Flooring covers the whole surface, so the question asks for the area. The area needs the length and the width, but the width is missing. Let w stand for the width in meters.
$P = 86 \text{ m}, \quad \ell = 28 \text{ m}, \quad w = \text{width in m}$
Write the perimeter rule for a rectangle and put in the values you know. This links the boundary line to the missing width.
$86 = 2(28 + w)$
Divide both sides by 2. The two sides are the same amount, so half of each is still the same amount. Dividing by 2 undoes the doubling, which leaves one length plus one width equal to 43 m.
$\frac{86}{2} = \frac{2(28 + w)}{2} \quad\text{so}\quad 43 = 28 + w$
Subtract 28 from both sides. Taking the same amount away from two equal amounts leaves them equal. Subtracting 28 undoes adding 28, so the width stands alone. The width is 15 m.
$43 - 28 = 28 + w - 28 \quad\text{so}\quad 15 = w$
Check the width in the perimeter rule before you use it. It gives back the 86 m boundary line, so the width is correct.
$2(28 + 15) = 2 \times 43 = 86$
Now find the area. Multiply the length by the width, the two perpendicular sides. Split 15 into 10 and 5 so you can multiply in parts.
$A = 28 \times 15 = (28 \times 10) + (28 \times 5) = 280 + 140 = 420 \text{ m}^2$

Answer

The court is $15$ m wide, so the crew needs $420$ square meters of flooring.

Now for the second skill: comparing two figures. A perimeter is a length and an area is a surface, so they have different units. You cannot say whether $20$ units is more or less than $24$ square units. So compare like with like: perimeter with perimeter, and area with area. That means you first find both measures for each figure.

The figure below shows two shapes on a coordinate grid. The horizontal axis is the $x$ -axis and the vertical axis is the $y$ -axis. In a point such as $(7, 1)$ , the first number is the $x$ -coordinate and the second is the $y$ -coordinate. Each grid step is $1$ unit, and every vertex (corner point) sits where two gridlines cross.

The left shape has vertices $(1, 1)$ , $(7, 1)$ , $(7, 5)$ and $(1, 5)$ .
The right shape has vertices $(9, 1)$ , $(14, 1)$ , $(14, 6)$ and $(9, 6)$ .

Do not judge a shape by how it looks. Use its vertices. Go around each shape from one vertex to the next. When only the $x$ -coordinate changes, the side is horizontal. When only the $y$ -coordinate changes, the side is vertical. In both shapes the sides go horizontal, vertical, horizontal, vertical. So every corner joins a horizontal side and a vertical side. Horizontal and vertical gridlines are perpendicular, so every corner is a right angle. Both shapes are rectangles.

Left rectangle:

(1, 1)

,

(7, 1)

,

(7, 5)

,

(1, 5)

. Right rectangle:

(9, 1)

,

(14, 1)

,

(14, 6)

,

(9, 6)

. Each grid step is

1

unit.

Worked example

Same distance around, different surface

Use the figure above, where each grid step is $1$ unit. The left rectangle has vertices $(1, 1)$ , $(7, 1)$ , $(7, 5)$ and $(1, 5)$ . The right rectangle has vertices $(9, 1)$ , $(14, 1)$ , $(14, 6)$ and $(9, 6)$ . Which rectangle has the greater perimeter? Which has the greater area?

Start with the left rectangle. From (1, 1) to (7, 1) only the x-coordinate changes, so that side is horizontal. From (7, 1) to (7, 5) only the y-coordinate changes, so that side is vertical. Each grid step is 1 unit, so each change is a length in units. The horizontal side is the base, or length. The vertical side is its perpendicular height, or width.
$\ell = 7 - 1 = 6, \quad w = 5 - 1 = 4$
Read the right rectangle the same way. From (9, 1) to (14, 1) the x-coordinate changes by 5. From (14, 1) to (14, 6) the y-coordinate changes by 5. Its length and width are equal, so this rectangle is a square. A square is still a rectangle, so the same two rules apply.
$\ell = 14 - 9 = 5, \quad w = 6 - 1 = 5$
Find both perimeters. Add the length and the width, then double, because you pass each of them twice on the way around. Both perimeters are 20 units.
$P_{\text{left}} = 2(6 + 4) = 2 \times 10 = 20, \quad P_{\text{right}} = 2(5 + 5) = 2 \times 10 = 20$
Find both areas. Multiply the length by the width. This counts the squares, 1 unit on each side, that fill each rectangle. The left one holds 24 square units and the right one holds 25.
$A_{\text{left}} = 6 \times 4 = 24, \quad A_{\text{right}} = 5 \times 5 = 25$
Compare like with like. The perimeters are equal. The areas are not equal: the right rectangle holds one more square unit.
$20 = 20, \quad 24 < 25$

Answer

Neither perimeter is greater, because both are $20$ units. The right rectangle, which is a square, has the greater area: $25$ square units compared with $24$ square units.

Here is a second pair. Use it to test an idea: does the rectangle with the longer perimeter always have the larger area?

In the figure below, each grid step is again $1$ unit, and every vertex sits where two gridlines cross.

The left shape has vertices $(1, 1)$ , $(11, 1)$ , $(11, 2)$ and $(1, 2)$ . Its horizontal sides are $11 - 1 = 10$ units long. Its vertical sides are $2 - 1 = 1$ unit long.
The right shape has vertices $(13, 1)$ , $(17, 1)$ , $(17, 5)$ and $(13, 5)$ . Its horizontal sides are $17 - 13 = 4$ units long. Its vertical sides are $5 - 1 = 4$ units long.

In both shapes, every corner joins a horizontal side and a vertical side, so every corner is a right angle. Both shapes are rectangles. The right one is also a square, because its four sides are equal.

Left rectangle:

(1, 1)

,

(11, 1)

,

(11, 2)

,

(1, 2)

. Right rectangle:

(13, 1)

,

(17, 1)

,

(17, 5)

,

(13, 5)

. Each grid step is

1

unit.

Takeaway. Perimeter is the distance around the whole edge of a rectangle: $P = 2(\ell + w)$ , in length units. Area is the surface inside: $A = \ell \times w$ , in square units. In both rules, $\ell$ and $w$ are the perpendicular sides in the same unit, and a square counts as a rectangle. In a word problem, first ask whether the story is about the edge or the surface. To compare two rectangles, find both measures for each one and compare like with like. Equal perimeters do not promise equal areas, and a longer perimeter does not promise a larger area.