Grade 4 · 4.MD.A.3

Finding a missing side from the area

Ljubica’s dog run has to cover 72 square feet, and the side along the fence is 8 feet. How long is the other side, and how much fence goes around it?

You need

Area of rectangles and squaresGrade 3 What division meansGrade 3

Where you are 15 in Area and perimeter

Measurement12

Area and perimeter15

▴ 4 above Perimeter Area of parallelograms and trapezoids

Finding a missing side from the area

Finding a missing side from the perimeter Perimeter versus area ▾ Show all 15 (6 more)

When you reach for it

Most of the time an area problem gives you the sides and asks for the area. You reach for this move when the problem runs the other way: you know how much space a rectangle must cover, and you know one side, and you need the other side.

Ljubica is planning a dog run. Her dog needs 72 square feet of ground, and one side has to be 8 feet long to fit along the yard fence. She cannot buy fencing until she knows the other side. Warrick has a swimming pool cover with its area printed on the label and wants to know how long it is. Both questions are about a missing side.

The second half of the lesson brings in perimeter, the distance around a shape (see Perimeter). A fence goes around a dog run, so the fence is a perimeter, while the grass inside is an area. Some problems give one and ask for the other.

The procedure

This lesson takes two days. Part 1 finds a missing side from the area. Part 2 finds a missing side from the perimeter, and goes back and forth between area and perimeter.

Part 1: the other side from the area

The area of a rectangle is its length times its width (see Area of rectangles and squares). The reason is the grid: a rectangle 8 feet by 9 feet holds 9 rows of 8 square feet, so 9 \times 8 = 72 squares, and its area is 72 square feet.

\text{area} = \text{length} \times \text{width}

Now hide one side. Ljubica knows the area, 72 square feet, and one side, 8 feet. Call the missing side w. The area fact still holds, so

8 \times w = 72.

This is a multiplication with a missing factor. A missing factor is found by division, because division asks exactly that question: 8 times what makes 72? In the grid, the 72 squares sit in rows of 8, and the number of rows is the missing side. Counting how many rows of 8 fit in 72 is the “how many groups” kind of division (see What division means).

Figure 172 square feet laid out in rows of 8. Each row is one foot deep, so the number of rows is the other side, and 72 \div 8 = 9.

w = 72 \div 8 = 9

So the other side is 9 feet. The same equation shape shows up when one amount is a number of times another, as in Times as many: “72 is 8 times what number?”

Two things are worth seeing here.

The units change. The area is in square feet, but the answer is in feet. Each row of the grid is a strip 1 foot deep, so counting rows counts feet of length. Square feet divided by feet gives feet.

The check is free. Multiply back: 8 \times 9 = 72. If the product is the area, the side is right. This works because division and multiplication undo each other.

When the area is large, divide the way you divide any large number by one digit (see Dividing by a single digit). An area model helps: split the area into pieces that each divide easily.

Sometimes the rectangle is a square and no side is given. Then both sides are the same unknown, so you look for a number that times itself makes the area.

Part 2: perimeter, and area and perimeter together

The perimeter of a rectangle is the length of all four sides added up. Opposite sides of a rectangle are equal, so a rectangle with length \ell and width w has

P = \ell + w + \ell + w.

To find a missing side from the perimeter (see Finding a missing side from the perimeter), notice that one length and one width together go halfway around. So half the perimeter is length plus width. Take away the side you know and the other side is left.

There is a second way that gives the same answer: take away both known sides from the perimeter, and what is left is the two equal unknown sides, so halve it. Both work because the perimeter is made of two copies of length plus width.

Area and perimeter can also be linked in one problem. If you know the area and one side, find the other side by division, then add all four sides for the perimeter. If you know the perimeter and one side, find the other side from half the perimeter, then multiply for the area.

One more fact: the area alone does not decide the perimeter. A rectangle of 24 square meters could be long and thin or close to a square, and the fence around it changes a lot. That is why a missing side problem always gives you one side as well as the area.

Worked examples

Examples 1 to 3 go with Part 1. Examples 4 to 7 go with Part 2.

Example 1

Ljubica’s dog run

Ljubica’s dog run must cover 72 square feet. One side is 8 feet. How long is the other side?

Solution

Let w be the other side in feet. Area is length times width, so

8 \times w = 72.

The missing factor is found by division: w = 72 \div 8 = 9. On the grid, 72 square feet in rows of 8 makes 9 rows, and each row is 1 foot deep, so the side is 9 feet.

Check by multiplying back: 8 \times 9 = 72. The other side is 9 feet.

Example 2

The pool cover

Warrick’s swimming pool cover has an area of 336 square feet. It is 8 feet wide. How long is the cover?

Solution

Let \ell be the length in feet. Then 8 \times \ell = 336, so \ell = 336 \div 8.

No times-table fact is that big, so split the area into two pieces that each divide by 8. Since 8 \times 40 = 320, cut the cover into a piece of 320 square feet and a piece of 16 square feet:

336 \div 8 = 320 \div 8 + 16 \div 8 = 40 + 2 = 42.

Splitting the area this way is allowed by the distributive property: 8 \times (40 + 2) is 8 \times 40 + 8 \times 2, which is 320 + 16. The picture shows it. The two pieces stand side by side, 8 feet tall, and their widths add to the whole length.

Figure 2336 split into 320 and 16, two areas that each divide easily by 8. The two widths, 40 feet and 2 feet, make the missing side, 42 feet.

The cover is 42 feet long. Check: 42 \times 8 = 320 + 16 = 336.

Example 3

A square dog run

Tarni wants a square dog run with an area of 64 square feet. How long is each side?

Solution

A square has four equal sides, so no side is given; both sides are the same unknown. Look for a number that times itself makes 64.

7 \times 7 = 49 is too small. 9 \times 9 = 81 is too big. 8 \times 8 = 64 is just right.

Each side is 8 feet. There is no other choice: a longer side makes a bigger square and a shorter side a smaller one, so only one side length gives exactly 64 square feet.

Example 4

A missing side from the fence

Tarni builds a rectangular dog run with 46 feet of fence all the way around. One side is 14 feet. How long is the other side?

Figure 3Tarni’s dog run. The four sides add to 46 feet, and opposite sides are equal.

Solution

Let w be the other side. The fence is the perimeter:

14 + w + 14 + w = 46.

One long side and one short side go halfway around, so 14 + w is half of 46, which is 23. Then w = 23 - 14 = 9.

The other way gives the same: take away both 14-foot sides, 46 - 28 = 18, and that is two equal short sides, so each is 18 \div 2 = 9.

The other side is 9 feet. Check: 14 + 9 + 14 + 9 = 46.

Example 5

From the area to the edging

Warrick’s second pool cover has an area of 180 square feet and is 9 feet wide. He wants to sew a rope trim all the way around its edge. How much rope does he need?

Solution

The rope goes around the edge, so it is a perimeter. But only one side is known, so first find the other side from the area: 9 \times \ell = 180, and

\ell = 180 \div 9 = 20.

This is easy to see because 9 \times 2 = 18, so 9 \times 20 = 180. The cover is 20 feet by 9 feet. Now add the four sides:

P = 20 + 9 + 20 + 9 = 58.

Warrick needs 58 feet of rope. The answer is in feet, not square feet, because rope is a length.

Example 6

From the fence to the grass

Ljubica fences a second dog run with 36 meters of fence. One side is 10 meters. She wants to cover the ground with grass. How much ground is that?

Solution

Half of the fence is one length plus one width: 36 \div 2 = 18. The other side is 18 - 10 = 8 meters. Check: 10 + 8 + 10 + 8 = 36.

The ground is the area:

10 \times 8 = 80.

The grass covers 80 square meters. The fence, 36 meters, is a length; the grass, 80 square meters, is an area. They measure different things and are never added together.

Example 7

Same area, different fence

Tarni has enough grass for 24 square meters and wants a rectangular dog run with sides that are whole meters. Which rectangles can she make, and which one needs the least fence?

Solution

The sides must multiply to 24. Try each first side and find the other by division, keeping only the ones that divide evenly:

Sides (meters)	Area (square meters)	Fence (meters)
1 by 24	24	1 + 24 + 1 + 24 = 50
2 by 12	24	2 + 12 + 2 + 12 = 28
3 by 8	24	3 + 8 + 3 + 8 = 22
4 by 6	24	4 + 6 + 4 + 6 = 20

A side of 5 does not work, since 24 \div 5 is not a whole number. A side of 6 gives 6 by 4, the same rectangle as 4 by 6 turned on its side, so the list is complete.

Figure 4Every rectangle here covers 24 square meters, but the fence around each one is a different length.

All four cover 24 square meters, but the fence runs from 20 meters up to 50. The 4 by 6 run needs the least fence, 20 meters. The closer a rectangle is to a square, the less edge it has for the same area.

Boundaries

The shape must be a rectangle. Area is length times width only for rectangles (squares included). For a triangle or an L shape, dividing the area by one side does not give another side.

The units must match. If the area is in square feet, the known side must be in feet, and then the answer is in feet. A side in yards with an area in square feet has to be changed first (see Converting square and cubic units). The answer is always a length, never square units.

The other side need not be a whole number. A rectangle of 30 square feet with a side of 7 feet has another side of 30 \div 7 feet. That division leaves a remainder, 4 \text{ R } 2, but the side is not “4 with 2 left over”; it is a length a little more than 4 feet, exactly 4\tfrac{2}{7} feet. In this lesson the numbers are chosen to divide evenly.

Area alone, or perimeter alone, does not fix a rectangle. You need one side as well. Example 7 shows four different rectangles with the same area. A missing side problem without a known side has many answers.

From a perimeter, the shape must have equal opposite sides. Halving the perimeter works because a rectangle’s sides come in equal pairs. For a shape with sides of four different lengths, you must know three of them.

Where this goes next

The same move works one step up. A box has a volume of length times width times height, so if you know the volume and two of the edges, dividing finds the third; that is how Volume of a box finds a missing edge. Comparing area and perimeter, as in Example 7, grows into Perimeter versus area, where the question is how far the two can pull apart.

Exercises

Part 1: the other side from the area

A rectangle has an area of 42 square feet. One side is 6 feet. Find the other side.
Write a multiplication equation with w for the missing side, then solve it: the area is 54 square meters and one side is 6 meters.
A rectangle covers 56 square meters and is 7 meters wide. How long is it?
A rectangle has an area of 150 square feet and one side of 5 feet. Find the other side.
A rectangle has an area of 252 square meters and one side of 7 meters. Find the other side. Split the area into two pieces first.
A rectangle has an area of 1{,}248 square feet and one side of 4 feet. Find the other side.
A square has an area of 100 square centimeters. How long is each side?
Warrick’s third pool cover is 288 square feet and 9 feet wide. How long is it?
A rectangle on a grid is made of 45 unit squares in 5 rows. How many squares are in each row, and what are the two side lengths in units?

Part 2: perimeter, and area and perimeter together

A rectangle has a perimeter of 34 feet. One side is 12 feet. Find the other side.
A rectangle has a perimeter of 50 meters. One side is 16 meters. Find the other side.
A square has a perimeter of 44 inches. How long is each side?
Tarni has 62 feet of fence for a rectangular dog run. One side is 20 feet. How long is the other side?
A rectangle has an area of 40 square feet and one side of 5 feet. Find its perimeter.
A rectangle has an area of 63 square meters and one side of 7 meters. Find its perimeter.
A rectangle has a perimeter of 28 feet and one side of 6 feet. Find its area.
A rectangle has a perimeter of 40 meters and one side of 13 meters. Find its area.
List every rectangle with whole-foot sides and an area of 18 square feet. Find the perimeter of each. Which needs the least fence?
Ljubica has 20 meters of fence for a rectangular dog run with whole-meter sides. List every rectangle she can make and its area. Which one gives her dog the most ground?
Warrick’s pool cover has an area of 240 square feet and is 8 feet wide. Trim for the edge costs 3 dollars a foot. How much does trim for the whole edge cost?
Tarni’s dog run sits against a wall, so only three sides need fence. The side along the wall is 15 meters and the area is 120 square meters. How much fence does she need?

Reasoning

Use rows of squares on a grid to explain why dividing the area of a rectangle by one side gives the other side.
Warrick says, “A rectangle with an area of 36 square feet always has a perimeter of 24 feet.” Is he right? Explain with two rectangles.
The area of a rectangle is 72 square feet and one side is 8 feet. Explain why the missing side is 9 feet and not 9 square feet.
Ljubica finds a missing side from the perimeter by halving the perimeter and taking away the known side. Explain why halving is allowed for a rectangle.

Answers

7 feet. 6 \times w = 42, so w = 42 \div 6 = 7. Check: 6 \times 7 = 42.
6 \times w = 54, so w = 54 \div 6 = 9 meters. Check: 6 \times 9 = 54.
8 meters. 56 \div 7 = 8, and 7 \times 8 = 56.
30 feet. 150 \div 5 = 30, since 5 \times 30 = 150.
36 meters. Split 252 into 210 + 42: 210 \div 7 = 30 and 42 \div 7 = 6, so 252 \div 7 = 36. Check: 7 \times 36 = 252.
312 feet. 1{,}248 \div 4 = 312, since 4 \times 300 = 1{,}200, 4 \times 12 = 48, and 1{,}200 + 48 = 1{,}248.
10 centimeters. 10 \times 10 = 100, and no other side length gives 100.
32 feet. 288 \div 9 = 32, since 9 \times 30 = 270 and 9 \times 2 = 18, and 270 + 18 = 288.
9 squares in each row. 45 \div 5 = 9, so the rectangle is 9 units by 5 units.
5 feet. Half the perimeter is 34 \div 2 = 17, and 17 - 12 = 5. Check: 12 + 5 + 12 + 5 = 34.
9 meters. 50 \div 2 = 25, and 25 - 16 = 9.
11 inches. A square has four equal sides, so 44 \div 4 = 11.
11 feet. 62 \div 2 = 31, and 31 - 20 = 11. Check: 20 + 11 + 20 + 11 = 62.
26 feet. The other side is 40 \div 5 = 8 feet, so P = 5 + 8 + 5 + 8 = 26.
32 meters. The other side is 63 \div 7 = 9 meters, so P = 7 + 9 + 7 + 9 = 32.
48 square feet. Half the perimeter is 14, so the other side is 14 - 6 = 8 feet, and the area is 6 \times 8 = 48.
91 square meters. Half the perimeter is 20, so the other side is 20 - 13 = 7 meters, and the area is 13 \times 7 = 91.
1 by 18 with perimeter 38 feet, 2 by 9 with perimeter 22 feet, and 3 by 6 with perimeter 18 feet. Sides of 4 and 5 do not divide 18 evenly. The 3 by 6 rectangle needs the least fence.
Length plus width is 20 \div 2 = 10, so the rectangles are 1 by 9 (area 9), 2 by 8 (area 16), 3 by 7 (area 21), 4 by 6 (area 24) and 5 by 5 (area 25), all in square meters. The 5 by 5 square gives the most ground, 25 square meters.
228 dollars. The length is 240 \div 8 = 30 feet, the edge is 30 + 8 + 30 + 8 = 76 feet, and 76 \times 3 = 228.
31 meters. The other side is 120 \div 15 = 8 meters, since 15 \times 8 = 120. The fence covers one 15-meter side and two 8-meter sides: 15 + 8 + 8 = 31.
The squares sit in rows, one row for each unit of one side, and each row holds as many squares as the other side is long. The area is the total number of squares. Dividing the total by the number in each row counts the rows, and the number of rows is the other side.
No. A 6 by 6 square has area 36 square feet and perimeter 24 feet, but a 4 by 9 rectangle also has area 36 square feet and its perimeter is 4 + 9 + 4 + 9 = 26 feet. The same area can come with different perimeters.
The 72 square feet sit in rows of 8, and each row is a strip 1 foot deep. Dividing 72 \div 8 = 9 counts rows, so it counts feet of the other side. A side is a length, measured in feet; square feet measure a surface.
A rectangle’s opposite sides are equal, so its perimeter is length plus width, twice over. Half of it is one length plus one width. Taking away the known side leaves the other.