Grade 3 · 3.MD.C.7D

Splitting a figure into rectangles

A tree house floor is shaped like an L, so it is not a rectangle and length times width does not fit it. How can its area still be found with the multiplication that works for rectangles?

You need

Area of rectangles and squaresGrade 3

Where you are 15 in Area and perimeter

Measurement12

Area and perimeter15

▴ 7 above Finding a missing side from the perimeter Perimeter versus area

Splitting a figure into rectangles

Area of triangles Area of any polygon by cutting it into triangles and rectangles ▾ Show all 15 (3 more)

When you reach for it

In Area of rectangles and squares, the area of a rectangle came from one multiplication: the side lengths say how many unit squares sit in each row and how many rows there are. That works because every row of a rectangle is the same length.

Many floors, yards and rooms are not rectangles, but they are close. Efia’s tree house floor is shaped like an L. Its rows are not all the same length, so no single product of two sides gives its area. Yet every corner of the floor is a square corner, and every side runs either straight across or straight up and down. A shape like that can always be cut into rectangles.

So this is the move to reach for when a flat shape has only square corners but is not one rectangle: an L shape, a T shape, a shape that steps down like stairs. Cut it into rectangles, find the area of each with length times width, and add. Counting the unit squares one by one, as in Measuring area with unit squares, would also work, and on a grid it is the check.

The procedure

The lesson runs over three days. Part 1 cuts a shape into rectangles and adds their areas, and says why adding is allowed. Part 2 finds the side lengths a plan leaves unmarked, which a cut usually needs. Part 3 cuts the same shape two different ways, checks both by counting squares, and meets one other route: a big rectangle with a corner taken away.

Part 1: cut into rectangles, then add

Here is Efia’s tree house floor, drawn on a grid of squares 1 foot on a side.

Figure 1One cut turns the L into two rectangles. Their areas, 24 and 6 square feet, add to the area of the floor.

The dashed line runs straight across from the inside corner of the L to the far side. It cuts the floor into two rectangles. Rectangle A along the bottom is 6 feet long and 4 feet wide, so it holds 4 rows of 6 square feet. Rectangle B on top is 3 feet long and 2 feet wide, so it holds 2 rows of 3:

4 \times 6 = 24 \qquad 2 \times 3 = 6

The whole floor is

24 + 6 = 30

square feet.

Why adding is allowed. Area counts unit squares. Every square of the floor lies in rectangle A or in rectangle B. None lies in both, because the cut only separates them. None is left out, because the two rectangles together are the whole floor. So counting A’s squares and then B’s squares counts every square of the floor exactly once. That is the fact this lesson rests on:

> When a shape is cut into pieces that do not overlap and leave no gaps, the area of the shape is > the sum of the areas of the pieces.

This is sometimes put as “area is additive”. It is the same idea as joining two groups of counters and adding to find how many there are in all: the squares are the counters.

The check. On a grid the squares can be counted. Row by row from the top, the floor has rows of 3, 3, 6, 6, 6, 6 squares, and 3 + 3 + 6 + 6 + 6 + 6 = 30. The count and the sum agree, as they must, since they count the same squares.

When the pieces share a whole side. Sometimes two rectangles meet along a whole side of the same length, and together they make one bigger rectangle. Two platforms each 4 feet deep, one 6 feet long and one 3 feet long, laid end to end, make a rectangle 4 feet deep and 9 feet long. Its area can be found either way:

4 \times 6 + 4 \times 3 = 24 + 12 = 36 \qquad 4 \times 9 = 36

So 4 \times 6 + 4 \times 3 = 4 \times 9. That is the The distributive property seen as area, and it is the same picture that the area model uses to break a hard product into easier ones. An L shape is what happens when the two pieces do not share a whole side: then they never join into one rectangle, and the two products have to be found and added separately.

Part 2: finding the sides the plan leaves out

A floor plan rarely marks the sides of the pieces. It marks the outside of the shape, and the cut needs lengths the plan does not show. With square corners those lengths are never lost.

Figure 2The plan marks four sides. The two at the inside corner are found from them by subtracting.

Look at this plan of an L. Every side runs either across or up and down. Walk along the top of the shape from left to right: first the top side, then the short side across at the inside corner. Those two together stretch across the whole shape, exactly as far as the bottom side does. So the short side across is the bottom side take away the top side:

8 - 5 = 3 \text{ feet}

The same goes up and down. On the right of the shape there are two up-and-down sides, the short one at the far right and the one at the inside corner. Together they rise as high as the whole left side. So the side at the inside corner is the left side take away the short right side:

7 - 3 = 4 \text{ feet}

In general, for a shape with square corners:

the sides that run across, measured along the top, add up to the sides that run across along the bottom (when the shape has no notch, all the top pieces together match the bottom);
the sides that run up and down on one side add up to the sides that run up and down on the other.

A missing side is then found by adding pieces, or by taking a known piece away from a whole. Once every side is known, the cut can be placed and each piece measured (Example 3).

A T shape works the same way, with one more thing to notice. In the courtyard below, the stem’s length is not marked, but the depth of the whole courtyard is, and so is the depth of the bar.

Figure 3The T is a bar and a stem. The stem’s length is not marked, but the whole depth is.

From the top of the bar to the end of the stem is 7 meters, and the bar takes 2 of them, so the stem is 7 - 2 = 5 meters long (Example 4).

Part 3: two cuts, one area

A shape with square corners can usually be cut in more than one way. The stepped floor of Ekua’s dance studio is drawn twice below, on a grid of squares 1 meter on a side, with two different cuts.

Figure 4Two different cuts of the same floor. Each square of the grid is 1 square meter, and both cuts count the same 37 squares.

The first cut runs across, making three strips. The second runs up and down, making three columns. The pieces are different, and so are the multiplications. But both sets of pieces cover the same floor with no gaps and no overlaps, so by the fact in Part 1 both sums give the same area (Example 5). Two cuts that give two different totals mean a piece was measured wrongly, or two pieces overlap, or part of the floor was left out.

That makes a second cut a check on the first, and on a grid a count of squares is a third. Off a grid, the second cut is the check that is always available.

Another way: the big rectangle with a corner missing. Some shapes look like a rectangle with a piece missing, and it can be quicker to see them that way. Efia’s L from Part 1 sits inside a square 6 feet on a side, with an empty corner 3 feet long and 2 feet wide.

Figure 5Another way to the same area. The floor is a 6 by 6 square with a 3 by 2 corner missing.

The square holds 6 \times 6 = 36 square feet. The empty corner would hold 2 \times 3 = 6 of them, and those 6 are not part of the floor. So the floor is 36 - 6 = 30 square feet, the same as before (Example 6). Taking away is allowed for the same reason adding was: the square’s squares are the floor’s squares and the corner’s squares, with none counted twice. This route returns in Area of composite figures and shaded regions, where shaded regions are found by subtracting.

Worked examples

Examples 1 and 2 go with Part 1, Examples 3 and 4 with Part 2, and Examples 5 to 7 with Part 3.

Example 1

Efia’s tree house floor

Efia’s tree house floor is an L made of a rectangle 6 feet long and 4 feet wide and a rectangle 3 feet long and 2 feet wide, set together with no overlap as in the figure. Find the area of the floor, and check it by counting squares on the grid.

Figure 6One cut turns the L into two rectangles. Their areas, 24 and 6 square feet, add to the area of the floor.

Solution

The cut along the dashed line gives the two rectangles. The bottom one holds 4 rows of 6 square feet, and the top one holds 2 rows of 3:

4 \times 6 = 24 \qquad 2 \times 3 = 6

The pieces do not overlap and together they are the whole floor, so their areas add:

24 + 6 = 30

The floor is 30 square feet.

Check by counting the grid squares a row at a time from the top: 3 + 3 + 6 + 6 + 6 + 6 = 30.

Example 2

Two platforms along one side

Kwasi builds a porch for the tree house from two rectangular platforms. Both are 7 feet long. One is 3 feet wide and the other 2 feet wide, and he lays them side by side so that their 7-foot sides touch along their whole length. Find the area of the porch two ways.

Solution

First way, piece by piece. The platforms are rectangles, and the porch is the two of them with no overlap, so

3 \times 7 + 2 \times 7 = 21 + 14 = 35

Second way, as one rectangle. The platforms share a whole 7-foot side, so together they make a rectangle 7 feet long and 3 + 2 = 5 feet wide:

5 \times 7 = 35

Both give 35 square feet. Rows of 7 show why: the first platform is 3 rows of 7 square feet, the second is 2 more rows of 7, and that is 5 rows of 7. So 3 \times 7 + 2 \times 7 = 5 \times 7, the The distributive property in a picture.

Example 3

The upper floor, with two sides missing

The plan of the tree house’s upper floor is an L with square corners. It marks the bottom side 8 feet, the left side 7 feet, the top side 5 feet and the short right side 3 feet. The two sides at the inside corner are not marked. Find them, and find the area of the floor.

Figure 7The plan marks four sides. The two at the inside corner are found from them by subtracting.

Solution

Every corner is square, so every side runs across or up and down.

Across: the top side and the short side at the inside corner together stretch as far as the bottom side. So the short side across is 8 - 5 = 3 feet.

Up and down: the short right side and the side at the inside corner together rise as high as the left side. So the side at the inside corner is 7 - 3 = 4 feet.

Now cut straight down from the inside corner. That makes a tall rectangle on the left, 5 feet across (the top side) and 7 feet tall (the left side), and a short rectangle on the right, 3 feet across and 3 feet tall (the short right side):

7 \times 5 = 35 \qquad 3 \times 3 = 9

They do not overlap and they make the whole floor, so the area is 35 + 9 = 44 square feet.

Check with the other cut, straight across from the inside corner. The bottom strip is 8 feet long and 3 feet tall, and the piece above it is 5 feet across and 4 feet tall (the side just found): 3 \times 8 + 4 \times 5 = 24 + 20 = 44. The same.

Example 4

The T-shaped courtyard

A courtyard is shaped like a T with square corners. The bar across the top is 9 meters long and 2 meters deep. The stem is 3 meters wide and hangs from the middle of the bar. From the top of the bar to the end of the stem is 7 meters. Find the area of the courtyard, and how far the bar reaches past the stem on each side.

Figure 8The T is a bar and a stem. The stem’s length is not marked, but the whole depth is.

Solution

Cut where the bar meets the stem, along the dashed line. That gives two rectangles.

The bar is 9 meters long and 2 meters deep, so it holds 2 rows of 9 square meters: 2 \times 9 = 18.

The stem’s length is not marked. The whole courtyard is 7 meters deep, and the bar takes up 2 meters of that, so the stem is 7 - 2 = 5 meters long. It is 3 meters wide, so it holds 5 \times 3 = 15 square meters.

The pieces do not overlap and they make the whole courtyard:

18 + 15 = 33

The courtyard is 33 square meters.

For the overhang: the bar is 9 meters and the stem takes 3 of them, leaving 9 - 3 = 6 meters. The stem is in the middle, so that 6 is shared equally between the two sides, 3 meters each. Check across the bar: 3 + 3 + 3 = 9.

Example 5

Ekua’s dance studio, cut two ways

Ekua’s dance studio floor is drawn on a grid of 1-meter squares. It is 8 meters across along the bottom 2 meters, 6 meters across along the next 2 meters, and 3 meters across along the top 3 meters, all lined up on the left. Find its area with a cut across, again with a cut up and down, and check by counting.

Figure 9Two different cuts of the same floor. Each square of the grid is 1 square meter, and both cuts count the same 37 squares.

Solution

Cut across (first drawing). The three strips are rectangles:

2 \times 8 + 2 \times 6 + 3 \times 3 = 16 + 12 + 9 = 37

Cut up and down (second drawing). Now the pieces are columns. The left column is 3 meters wide and runs the full 2 + 2 + 3 = 7 meters up. The middle column is 6 - 3 = 3 meters wide and covers the bottom two strips, 2 + 2 = 4 meters up. The right column is 8 - 6 = 2 meters wide and only 2 meters up:

7 \times 3 + 4 \times 3 + 2 \times 2 = 21 + 12 + 4 = 37

Both cuts give 37 square meters. The middle column needed a side the drawing does not mark, found by subtracting, as in Part 2.

Check by counting rows of squares from the top: three rows of 3, two rows of 6, two rows of 8. That is 3 + 3 + 3 + 6 + 6 + 8 + 8 = 37.

Example 6

The tree house floor, another way

Find the area of Efia’s floor from Example 1 by seeing it as a square with a corner missing.

Figure 10Another way to the same area. The floor is a 6 by 6 square with a 3 by 2 corner missing.

Solution

The floor is 6 feet along the bottom and 2 + 4 = 6 feet up the left side, so it fits exactly in a square 6 feet on each side. The square holds

6 \times 6 = 36

square feet. The part of the square that is not floor is the empty top right corner, 3 feet long (6 - 3) and 2 feet tall, which would hold 2 \times 3 = 6 square feet. Take it away:

36 - 6 = 30

The floor is 30 square feet, the same answer as Example 1. Adding the pieces and taking away the missing corner are two ways of counting the same squares.

Example 7

Planning a courtyard of a given size

Efia is planning an L-shaped courtyard of 48 square meters. One part is a rectangle 7 meters long and 4 meters wide. The rest is a second rectangle, 4 meters wide, set beside the first with no overlap. How long must the second rectangle be?

Solution

The two pieces add to the whole courtyard, so the second piece is what is left after the first:

4 \times 7 = 28 \qquad 48 - 28 = 20

The second rectangle must be 20 square meters. It is 4 meters wide, so picture its square meters in rows of 4, each row 1 meter long. How many rows of 4 make 20? Count in fours: 4, 8, 12, 16, 20. That is 5 rows, so 20 \div 4 = 5, and the second rectangle is 5 meters long. This backward step is Finding a missing side from the area.

Check: 5 \times 4 = 20 and 28 + 20 = 48.

Boundaries

Square corners only. Every piece here is a rectangle, and every unmarked side is found by adding or subtracting because each side runs straight across or straight up and down. A shape with a slanted side cannot be cut into rectangles alone; shapes like that need triangles, in Area of any polygon by cutting it into triangles and rectangles.

No overlap and no gaps. The sum of the pieces is the area only when the pieces fit together exactly. Two pieces that overlap count the shared squares twice (Exercise 16), and a cut that leaves part of the shape out of every piece counts too few.

One unit throughout. Every length must be in the same unit before multiplying, and every piece’s area is then in the same square unit, so the areas can be added. An area is never added to a length.

Whole-number sides. The pieces in this lesson have whole-number sides, so each holds a whole number of unit squares. The same cutting works for sides like \displaystyle 2\frac{1}{2} feet once rectangles with fraction sides are measured, in A fraction times a fraction.

A notch changes the side rule. The rule that the top pieces add to the bottom holds for an L, a T or a staircase. A shape with a notch cut into its top edge, like a U, has top sides that skip the notch, so the sides of the notch itself must also be counted. Draw it, and walk along each edge.

Where this goes next

The same cut-and-add move finds the area of any shape that can be split into pieces of known area. In Area of composite figures and shaded regions, the pieces can also be triangles and parts of circles, and the “big shape with a part missing” route of Example 6 is used for shaded regions. In Area of any polygon by cutting it into triangles and rectangles, any polygon, slanted sides and all, is cut into rectangles and triangles. And the same reasoning, one level up, finds the volume of a solid built from boxes, in Composite solids and density.

The link to multiplication runs the other way too. When two rectangles share a whole side, as in Example 2, the sum of their areas is one product, which is the The distributive property. That is the picture behind the area model.

The edge of these same shapes, rather than their inside, is their Perimeter, and Perimeter versus area puts the two measures side by side.

Exercises

Exercises 1 to 5 go with Part 1, 6 to 10 with Part 2, and 11 to 15 with Part 3. The reasoning group draws on all three.

Part 1: cut into rectangles, then add

A shape is made of a rectangle 5 feet long and 3 feet wide and a rectangle 2 feet long and 4 feet wide, set together with no overlap. What is its area?
Kwasi’s tree house floor is an L made of a rectangle 7 feet by 5 feet and a rectangle 2 feet by 3 feet, with no overlap. What is the area of the floor?
A courtyard is made of three rectangles that do not overlap: 6 meters by 5 meters, 4 meters by 2 meters, and 3 meters by 3 meters. What is the area of the courtyard?
Two platforms are each 6 feet long. One is 3 feet wide and the other 5 feet wide, and they touch along their whole 6-foot sides. Find the area of the two together in two ways.

The figure shows shapes A and B, for Exercises 5 and 11. Each grid square is 1 square centimeter.

Figure 11Shapes A and B, for Exercises 5 and 11.

Cut shape A into two rectangles and add their areas. Then check by counting the squares.

Part 2: finding the sides the plan leaves out

An L-shaped floor has square corners. Its bottom side is 9 feet, its left side 6 feet, its top side 4 feet and its short right side 2 feet. Find the two sides at the inside corner, and the area of the floor.
Efia’s tree house floor plan is an L with square corners. The bottom is 10 feet, the left side 8 feet, the top 4 feet and the short right side 3 feet. Find the two unmarked sides and the area.
A T-shaped courtyard has square corners. The bar across the top is 8 meters long and 2 meters deep. The stem is 2 meters wide and hangs from the middle of the bar. From the top of the bar to the end of the stem is 6 meters. How long is the stem, how far does the bar reach past the stem on each side, and what is the area?
A stepped floor has square corners. Its bottom side is 9 meters and its left side 6 meters. From the bottom right corner its edge goes up 2 meters, left 3 meters, up 2 meters, left 3 meters, then up to the top, then left along the top to the left side. How long are the last side going up and the top side? What is the area of the floor?
An L-shaped floor is 53 square feet. One part is a rectangle 7 feet by 5 feet. The other part is a rectangle 3 feet wide. How long is the other part?

Part 3: two cuts, one area

Find the area of shape B with a cut up and down, then with a cut across. Check by counting the squares.
Find the area of the floor in Exercise 6 another way: as a rectangle with a corner missing.
The hallway outside Ekua’s dance studio has an L-shaped floor with square corners. Its bottom is 8 meters, its left side 6 meters, its top 3 meters and its short right side 4 meters. Find the two unmarked sides, then the area with a cut up and down and again with a cut across.
A T-shaped courtyard has a bar 7 meters long and 2 meters deep, and a stem 3 meters wide hanging from its middle. The courtyard is 5 meters from the top of the bar to the end of the stem. Find its area as a bar and a stem, and again as three columns side by side.
One can of floor varnish covers 9 square meters. How many cans cover the stepped floor in Exercise 9?

Reasoning

Efia’s floor from Example 1 holds the 6 by 4 rectangle along the bottom and also the 3 by 6 strip up the left side. Their areas, 24 and 18, add to 42, but the floor is 30 square feet. Explain why these two pieces do not give the area by adding, and how many squares lie in both of them.
Explain why any L shape with square corners can be cut into two rectangles in two different ways.
Can two L-shaped floors with different side lengths have the same area? Give an example.
Draw two rectangles that show 3 \times 5 + 3 \times 4 = 3 \times 9, and explain the picture.
A plan shows an L with square corners and marks only the bottom side and the top side. Explain why the short side across at the inside corner can be found, and why the area still cannot.

Answers

23 square feet. The pieces are 3 \times 5 = 15 and 4 \times 2 = 8, and 15 + 8 = 23.
41 square feet. The pieces are 5 \times 7 = 35 and 3 \times 2 = 6, and 35 + 6 = 41.
47 square meters. The pieces are 5 \times 6 = 30, 2 \times 4 = 8 and 3 \times 3 = 9, and 30 + 8 + 9 = 47.
48 square feet both ways. Piece by piece, 3 \times 6 + 5 \times 6 = 18 + 30 = 48. As one rectangle 6 feet by 3 + 5 = 8 feet, 8 \times 6 = 48.
20 square centimeters. Cut across: the bottom strip is 2 \times 7 = 14 and the block on top is 2 \times 3 = 6, and 14 + 6 = 20. Counting rows from the top: 3 + 3 + 7 + 7 = 20.
The side across at the inside corner is 9 - 4 = 5 feet, and the side up and down there is 6 - 2 = 4 feet. Area 34 square feet: cut down from the inside corner, 6 \times 4 = 24 and 2 \times 5 = 10, and 24 + 10 = 34.
The side across is 10 - 4 = 6 feet and the side up and down is 8 - 3 = 5 feet. Area 50 square feet: the left piece is 8 \times 4 = 32 and the right piece 3 \times 6 = 18, and 32 + 18 = 50.
The stem is 6 - 2 = 4 meters long. The bar reaches 3 meters past the stem on each side, since 8 - 2 = 6 shared equally is 3 and 3. Area 24 square meters: the bar is 2 \times 8 = 16, the stem 4 \times 2 = 8, and 16 + 8 = 24.
The last side going up is 6 - 2 - 2 = 2 meters and the top is 9 - 3 - 3 = 3 meters. Area 36 square meters: cut across into strips 2 meters tall, 9, 6 and 3 meters long, so 2 \times 9 + 2 \times 6 + 2 \times 3 = 18 + 12 + 6 = 36.
6 feet. The first part is 5 \times 7 = 35, so the other part is 53 - 35 = 18 square feet. Rows of 3: \square \times 3 = 18, and 18 \div 3 = 6. Check: 35 + 18 = 53.
18 square centimeters. Up and down: columns 2 wide and 5, 3 and 1 tall, so 5 \times 2 + 3 \times 2 + 1 \times 2 = 10 + 6 + 2 = 18. Across: strips of 6, 4 and 2 squares long, 1, 2 and 2 tall, so 1 \times 6 + 2 \times 4 + 2 \times 2 = 6 + 8 + 4 = 18. Counting rows from the top: 2 + 2 + 4 + 4 + 6 = 18.
34 square feet. The floor fits in a rectangle 9 feet by 6 feet, which is 6 \times 9 = 54. The missing top right corner is 5 feet across and 4 feet tall, which is 4 \times 5 = 20. And 54 - 20 = 34, as in Exercise 6.
The side across at the inside corner is 8 - 3 = 5 meters and the side up and down is 6 - 4 = 2 meters. Area 38 square meters both ways. Up and down: 6 \times 3 + 4 \times 5 = 18 + 20 = 38. Across: 4 \times 8 + 2 \times 3 = 32 + 6 = 38.
23 square meters. The stem is 5 - 2 = 3 meters long. Bar and stem: 2 \times 7 + 3 \times 3 = 14 + 9 = 23. As columns: the bar reaches (7 - 3) \div 2 = 2 meters past the stem on each side, so the columns are 2 \times 2 = 4, 5 \times 3 = 15 and 2 \times 2 = 4, and 4 + 15 + 4 = 23.
4 cans. The floor is 36 square meters (Exercise 9), and 36 \div 9 = 4.
12 squares. The two pieces overlap: the bottom left corner of the floor, 3 feet long and 4 feet tall, lies in both the 6 by 4 rectangle and the 3 by 6 strip, so its 4 \times 3 = 12 squares are counted twice in 42. Taking them away once gives 42 - 12 = 30, the area of the floor. Adding gives the area only for pieces that do not overlap.
An L has exactly one inside corner. Two sides meet there, one running across and one up and down. Carrying the side that runs across straight on to the far edge cuts the L into two rectangles; carrying the side that runs up and down straight on to the far edge cuts it into two different rectangles. Both cuts give pieces with square corners on every side, so both are rectangles.
Yes. Efia’s floor is a 6 by 4 rectangle with a 3 by 2 block on top, 24 + 6 = 30 square feet, with bottom 6 feet and left side 6 feet. An L made of a rectangle 7 feet long and 4 feet wide with a block 2 feet long and 1 foot wide on top at the left has 4 \times 7 + 1 \times 2 = 28 + 2 = 30 square feet too, but its bottom is 7 feet and its left side 5 feet. Same area, different sides.
Draw a rectangle 3 units tall and 5 long, and beside it a rectangle 3 tall and 4 long, touching along their 3-unit sides. Their areas are 3 \times 5 = 15 and 3 \times 4 = 12. Together they make one rectangle 3 tall and 5 + 4 = 9 long, with area 3 \times 9 = 27, and 15 + 12 = 27. The same squares are counted both ways.
The top side and the short side across together stretch as far as the bottom side, so the short side is the bottom take away the top. But nothing up and down is marked, so the heights of the pieces are unknown, and an L with these two sides could be short or tall. Without a height the area cannot be found.