Decompose Composite Figures and Find Their Area | Grade 6 Math | LevelKite

You can already find the area of a rectangle, a triangle, a parallelogram and a trapezoid. Many figures are none of these. A composite figure is a figure made from two or more simpler shapes that are joined along their sides and do not overlap.

The area of a figure is the amount of surface inside it. You measure it by how many unit squares, including parts of squares, cover the figure exactly. Every figure in this lesson is drawn on a coordinate grid. The horizontal axis is the $x$ -axis and the vertical axis is the $y$ -axis. One step on the grid is $1$ unit of length, so one grid square has an area of $1$ square unit.

Counting squares one by one is slow, and a slanted side cuts through squares. There is a better way. To decompose a figure means to cut it into smaller pieces. A cut does not add any area and does not take any away. It only sorts the same area into pieces. So when you know the area of every piece, you can add those areas to get the area of the whole figure.

Key concept

Decompose, then add

The area of a composite figure equals the sum of the areas of its pieces. This is true only when the pieces cover the whole figure, stay inside it and do not overlap. Pieces may touch only along a cut.

Choose cuts that make rectangles and triangles, because you already know their area formulas. A square is a rectangle, so a square piece works too. In the formula, $A$ stands for area in square units. Add one area for every piece.

A_{\text{figure}} = A_{\text{piece 1}} + A_{\text{piece 2}} + \cdots

A vertex is a corner point of a figure. The plural is vertices. Vertices let you measure exactly, so you never have to trust how a figure looks.

A side is horizontal when its two end points have the same $y$ -coordinate. Its length is the larger $x$ -coordinate minus the smaller one.
A side is vertical when its two end points have the same $x$ -coordinate. Its length is the larger $y$ -coordinate minus the smaller one.
Two sides are perpendicular when they meet at a right angle. A horizontal side and a vertical side that meet are always perpendicular.

Let $b$ be the length of a base and $h$ be the height, both in units. The height must be perpendicular to the base. You already know two formulas that use them:

Rectangle: $A = b \times h$ , where $b$ and $h$ are the lengths of two sides that meet at a right angle.
Triangle: $A = \frac{1}{2} \times b \times h$ , where $h$ is the distance from the opposite vertex to the line through the base, measured perpendicular to the base.

The whole pentagon before any cut, with vertices

(1, 1)

,

(7, 1)

,

(7, 4)

,

(4, 7)

and

(1, 4)

.

Worked example

Decompose a pentagon

The figure above is a pentagon, a five-sided figure. Its vertices are $(1, 1)$ , $(7, 1)$ , $(7, 4)$ , $(4, 7)$ and $(1, 4)$ . Decompose it into one rectangle and one triangle.

Sort the five sides by reading the vertices. The side from (1, 1) to (7, 1) is horizontal, because both y-coordinates are 1. The sides from (7, 1) to (7, 4) and from (1, 4) to (1, 1) are vertical, because the x-coordinate stays the same along each one. The two sides that meet at (4, 7) are slanted, because both coordinates change along them.
Choose a cut that separates the slanted sides from the rest. The slanted sides begin at the vertices (1, 4) and (7, 4). Join these two vertices. Both have y-coordinate 4, so the cut is horizontal.
Check the lower piece. Its vertices are (1, 1), (7, 1), (7, 4) and (1, 4). Going around it, the sides are horizontal, vertical, horizontal, vertical. At every corner a horizontal side meets a vertical side, so every corner is a right angle. A four-sided figure with four right angles is a rectangle.
Check the upper piece. Its vertices are (1, 4), (7, 4) and (4, 7). It has three sides: the cut and the two slanted sides. The vertex (4, 7) has y-coordinate 7, so it is not on the cut. The piece is a triangle.
Check the two pieces together. Every point of the rectangle has a y-coordinate from 1 to 4. Every point of the triangle has a y-coordinate from 4 to 7. So the pieces touch only along the cut and do not overlap. Together they have the same outline as the pentagon, so nothing is missing and nothing is added.

Answer

The pentagon decomposes into a rectangle with vertices $(1, 1)$ , $(7, 1)$ , $(7, 4)$ , $(1, 4)$ and a triangle with vertices $(1, 4)$ , $(7, 4)$ , $(4, 7)$ .

The same pentagon after the cut from

(1, 4)

to

(7, 4)

, with the rectangle below the cut and the triangle above it.

Worked example

Find the area of the pentagon

The figure above shows the pentagon from the first example as two pieces. The rectangle has vertices $(1, 1)$ , $(7, 1)$ , $(7, 4)$ , $(1, 4)$ . The triangle has vertices $(1, 4)$ , $(7, 4)$ , $(4, 7)$ . Find the area of the pentagon.

Measure the rectangle. Take the horizontal side from (1, 1) to (7, 1) as the base and subtract the x-coordinates. Take the vertical side from (7, 1) to (7, 4) as the height and subtract the y-coordinates. These two sides meet at a right angle, so the height is perpendicular to the base. Both lengths are in units.
$b = 7 - 1 = 6, \quad h = 4 - 1 = 3$
Multiply the base by the height to get the rectangle's area in square units. You can check this on the grid: the rectangle holds 3 rows of 6 grid squares.
$A_{\text{rectangle}} = 6 \times 3 = 18$
Measure the triangle. Use the cut from (1, 4) to (7, 4) as the base, and subtract the x-coordinates. The base is horizontal, so the height must be vertical. Go straight down from the opposite vertex (4, 7) to the point (4, 4) on the base. Subtract the y-coordinates of these two points to get the height.
$b = 7 - 1 = 6, \quad h = 7 - 4 = 3$
Use the triangle formula. Multiply the base by the height first, then take half.
$A_{\text{triangle}} = \frac{1}{2} \times 6 \times 3 = \frac{1}{2} \times 18 = 9$
Add the two areas. This is valid because the two pieces cover the whole pentagon and do not overlap.
$A_{\text{pentagon}} = 18 + 9 = 27$

Answer

The area of the pentagon is $27$ square units.

In the pentagon, the cut joined the two vertices where the slanted sides begin. Other figures need more than one cut, and you often have a choice.

Two kinds of vertex are good places to start a cut: one where a slanted side begins, and one where the outline bends inward. From that vertex, try a cut along a gridline, straight across or straight up and down, until you reach another side of the figure. Then check each piece. If a piece is not yet a rectangle or a triangle, cut again.

Cuts along gridlines help because they give each piece a horizontal or vertical side. You can find the length of such a side by subtracting coordinates.

The next figure is a hexagon, a six-sided figure. Two horizontal cuts split it into three pieces.

A hexagon with vertices

(1, 1)

,

(8, 1)

,

(8, 3)

,

(4, 3)

,

(4, 6)

and

(1, 8)

, cut into three pieces from

(4, 3)

to

(1, 3)

and from

(4, 6)

to

(1, 6)

.

Worked example

A figure with three pieces

The figure above is a hexagon with vertices $(1, 1)$ , $(8, 1)$ , $(8, 3)$ , $(4, 3)$ , $(4, 6)$ and $(1, 8)$ . Decompose it into rectangles and triangles. Then find its area.

Choose the cuts. The outline bends inward at (4, 3), and the slanted side begins at (4, 6). Cut horizontally from (4, 3) to (1, 3) and from (4, 6) to (1, 6). Both cuts end on the vertical side from (1, 1) to (1, 8). The figure shows the three pieces that these cuts make.
Bottom piece: its vertices are (1, 1), (8, 1), (8, 3) and (1, 3). At every corner a horizontal side meets a vertical side, so it is a rectangle. Its base is the difference of the x-coordinates 8 and 1. Its height is the difference of the y-coordinates 3 and 1.
$A_{\text{bottom}} = (8 - 1) \times (3 - 1) = 7 \times 2 = 14$
Middle piece: its vertices are (1, 3), (4, 3), (4, 6) and (1, 6). Its corners are right angles for the same reason. Its base, from x-coordinate 1 to 4, and its height, from y-coordinate 3 to 6, are both 3 units. So it is a square. A square is a rectangle, so the same formula works.
$A_{\text{middle}} = (4 - 1) \times (6 - 3) = 3 \times 3 = 9$
Top piece: its vertices are (1, 6), (4, 6) and (1, 8). It has three sides, so it is a triangle. Use the cut from (1, 6) to (4, 6) as the base. The side from (1, 6) to (1, 8) is vertical, so it is perpendicular to the base. That side is the height.
$b = 4 - 1 = 3, \quad h = 8 - 6 = 2$
Use the triangle formula. Multiply the base by the height, then take half.
$A_{\text{top}} = \frac{1}{2} \times 3 \times 2 = \frac{1}{2} \times 6 = 3$
Add the three areas. The pieces cover the whole hexagon and touch only along the cuts, so no part is counted twice.
$A_{\text{hexagon}} = 14 + 9 + 3 = 26$

Answer

The hexagon decomposes into two rectangles, one of which is a square, and one triangle. Its area is $26$ square units.

The tall rectangle is not wrong by itself. It only needs a partner that stops at its edge. Cut the hexagon vertically from $(4, 3)$ down to $(4, 1)$ , and keep the cut from $(4, 6)$ to $(1, 6)$ . In each of the two new pieces, a horizontal side meets a vertical side at every corner, so both are rectangles. The pieces are now:

a tall rectangle with vertices $(1, 1)$ , $(4, 1)$ , $(4, 6)$ , $(1, 6)$ : base $4 - 1 = 3$ , height $6 - 1 = 5$ , area $3 \times 5 = 15$
a smaller rectangle with vertices $(4, 1)$ , $(8, 1)$ , $(8, 3)$ , $(4, 3)$ : base $8 - 4 = 4$ , height $3 - 1 = 2$ , area $4 \times 2 = 8$
the same triangle as before, with area $3$

The total is $15 + 8 + 3 = 26$ square units, the same as before. The expressions $14 + 9 + 3$ and $15 + 8 + 3$ are equal because each one counts every part of the hexagon exactly once. Only the grouping changes. This gives you a check: two ways of cutting the same figure, each with no gaps and no overlaps, must give the same area.

Takeaway: To find the area of a composite figure, decompose it into rectangles and triangles. Use the vertices to find each base and its perpendicular height. Find the area of each piece, then add. The sum is the area of the whole figure as long as the pieces cover all of it and do not overlap.

Next, you will use this idea to find the area between two figures, solve word problems and spot mistakes in worked solutions.