Subtract Areas and Check Composite Area Solutions | Grade 6 Math | LevelKite

You already know how to find the area of a composite figure. That is a figure built from simpler shapes. You split it into parts and add the areas of the parts.

Some figures are easier to see another way: as a big shape with a piece taken out, like a wall with a window in it. Call the big shape the outer figure and the piece inside it the inner figure.

The area between two figures is the area of the region that is inside the outer figure but outside the inner figure. It is what is left of the outer figure after the inner figure is taken away. In this lesson you will find areas like this, use them in word problems, and learn to spot mistakes in a solution.

Here is why subtraction finds the area between. Suppose the inner figure lies completely inside the outer figure. Then the outer figure is made of exactly two parts that do not overlap: the inner figure and the region between. The areas of those two parts add up to the whole:

$\text{outer area} = \text{inner area} + \text{area between}$

Try it with numbers. Area tells you how many unit squares it takes to cover a region exactly. A unit square is $1$ unit on each side. Suppose the outer figure covers $20$ unit squares and the inner figure covers $6$ of them. Let $b$ be the area between, in unit squares. Then $6 + b = 20$ . Subtracting $6$ undoes adding $6$ . Doing it on both sides keeps the two sides equal, so $b = 20 - 6 = 14$ . The region between covers $14$ unit squares.

Key concept

Area between two figures

When an inner figure lies completely inside an outer figure, the area between them is the outer area minus the inner area. Find each area first, in the same square units, and then subtract. If part of the inner figure sticks out past the outer figure, this rule does not apply.

Under the same condition you can use the add-back check: the area between plus the inner area must equal the outer area.

\text{area between} = \text{outer area} - \text{inner area}

Outer rectangle with corners

(1, 1)

,

(11, 1)

,

(11, 7)

,

(1, 7)

and inner rectangle with corners

(3, 2)

,

(7, 2)

,

(7, 5)

,

(3, 5)

.

Worked example

Area between two rectangles on a grid

The figure shows two rectangles on a coordinate grid. The $x$ -axis runs across, the $y$ -axis runs up, and each grid square is $1$ unit on a side. The outer rectangle has corners $(1, 1)$ , $(11, 1)$ , $(11, 7)$ and $(1, 7)$ . The inner rectangle has corners $(3, 2)$ , $(7, 2)$ , $(7, 5)$ and $(3, 5)$ . Find the area between the two rectangles.

Check that subtracting is allowed. The outer rectangle spans x from 1 to 11 and y from 1 to 7. The inner rectangle spans x from 3 to 7 and y from 2 to 5, which stays inside those ranges. So the inner rectangle lies completely inside the outer one.
Read the outer rectangle's side lengths from its corners. The side from (1, 1) to (11, 1) is horizontal, because both y-coordinates are 1. Its length is the difference of the x-coordinates. The side from (11, 1) to (11, 7) is vertical, because both x-coordinates are 11. Its length is the difference of the y-coordinates.
$11 - 1 = 10 \quad \text{and} \quad 7 - 1 = 6$
A horizontal side and a vertical side meet at a right angle, so 10 and 6 are the length and width of the rectangle. Multiply them to get the outer area in square units.
$10 \times 6 = 60$
Do the same for the inner rectangle. Its horizontal side runs from (3, 2) to (7, 2) and its vertical side runs from (7, 2) to (7, 5). Write each length as a difference of coordinates, then multiply.
$(7 - 3) \times (5 - 2) = 4 \times 3 = 12$
Subtract the inner area from the outer area. The region between is what is left of the outer rectangle after the inner rectangle is taken away.
$60 - 12 = 48$
Use the add-back check. The region between and the inner rectangle together must make up the whole outer rectangle.
$48 + 12 = 60$

Answer

The area between the two rectangles is $48$ square units.

In a word problem, the story tells you which areas to combine. Make a plan before you calculate:

Find the whole figure. If it is built from parts that do not overlap, add the areas of the parts.
Find what is left out, such as a window, a pond or a hole. If it lies completely inside the whole figure, subtract its area.
Find every length you need. Sometimes the story does not give a length directly, and you must work it out.
Keep the units. Lengths in inches give areas in square inches. One square inch is the area of a square that is $1$ inch on each side.

A quick sketch helps you see which shape is inside which.

Worked example

A border around a photo

A rectangular photo is $8$ inches long and $5$ inches wide. It is glued onto a rectangular card so that a border of card $1$ inch wide shows on every side of the photo. What is the area of the border?

Plan. The border is the region between two rectangles. The outer figure is the card. The inner figure is the photo, which lies completely inside the card. So the border's area is the card's area minus the photo's area.
Work out the card's side lengths. The border is 1 inch wide on every side. So the card is 1 inch longer than the photo at both ends, and 1 inch wider on both sides. Add 1 inch twice to each measurement.
$8 + 1 + 1 = 10 \quad \text{and} \quad 5 + 1 + 1 = 7$
Multiply length by width for each rectangle. Inches times inches gives square inches.
$10 \times 7 = 70 \quad \text{and} \quad 8 \times 5 = 40$
Subtract the photo's area from the card's area. What is left is the border.
$70 - 40 = 30$
Use the add-back check. The border and the photo together cover the whole card.
$30 + 40 = 70$

Answer

The area of the border is $30$ square inches.

Shed wall for the next example, on a grid where each step is

1

foot: a rectangle with a triangle on top, and a window (the smaller rectangle) inside.

Worked example

Painting a shed wall

The figure shows the front wall of a shed on a coordinate grid. One grid step is $1$ foot on both the $x$ -axis (across) and the $y$ -axis (up). The wall is a rectangle with corners $(1, 1)$ , $(9, 1)$ , $(9, 5)$ , $(1, 5)$ , with a triangle on top that has corners $(1, 5)$ , $(9, 5)$ , $(5, 8)$ . A window has corners $(3, 2)$ , $(7, 2)$ , $(7, 4)$ , $(3, 4)$ . Kai paints the wall but not the window. What area does Kai paint?

Plan. The rectangle and the triangle share only the side from (1, 5) to (9, 5), so they do not overlap. Add their areas to get the whole wall. The rectangle part spans x from 1 to 9 and y from 1 to 5. The window spans x from 3 to 7 and y from 2 to 4, so it lies completely inside the wall. Subtract its area.
Rectangle part. Its side from (1, 1) to (9, 1) is horizontal and its side from (9, 1) to (9, 5) is vertical, so they meet at a right angle. Write each length as a difference of coordinates, then multiply.
$(9 - 1) \times (5 - 1) = 8 \times 4 = 32$
Triangle part. Use the horizontal side from (1, 5) to (9, 5) as the base. The height goes from the top corner (5, 8) straight down to (5, 5) on the base. That segment is vertical, so it is perpendicular to the base: it meets the base at a right angle. The area is half of the base times this height.
$\frac{1}{2} \times (9 - 1) \times (8 - 5) = \frac{1}{2} \times 8 \times 3 = \frac{1}{2} \times 24 = 12$
Add the two parts to get the area of the whole wall in square feet.
$32 + 12 = 44$
Window. Its sides lie along grid lines, so it is a rectangle. Its horizontal side runs from (3, 2) to (7, 2) and its vertical side runs from (7, 2) to (7, 4).
$(7 - 3) \times (4 - 2) = 4 \times 2 = 8$
Subtract the window's area from the whole wall, because Kai does not paint the window.
$44 - 8 = 36$

Answer

Kai paints $36$ square feet of the wall. The add-back check agrees: $36 + 8 = 44$ , the area of the whole wall.

Sometimes your job is to find a mistake in a solution. Do not just look at the final number. Go through the work one line at a time and ask three questions:

Plan: Is each part added or subtracted correctly? Parts that join without overlapping are added. A piece taken out from inside the figure is subtracted.
Measurements: Does every length match the figure? Is each height perpendicular to its base?
Formulas and arithmetic: Is the right area formula used for each shape, and is each calculation correct?

The first line that fails one of these questions is where the mistake starts. Later lines may carry the wrong value forward, so recalculate them.

Try the three questions on Dev's work. Dev wanted the area between the rectangle and the triangle in the figure below. On the grid, the $x$ -axis runs across, the $y$ -axis runs up, and each grid square is $1$ unit on a side. The rectangle has corners $(1, 1)$ , $(9, 1)$ , $(9, 7)$ and $(1, 7)$ . The triangle has corners $(2, 2)$ , $(8, 2)$ and $(4, 6)$ . Dev's work is in the table under the figure.

Dev's figure: a rectangle with corners

(1, 1)

,

(9, 1)

,

(9, 7)

,

(1, 7)

and a triangle inside with corners

(2, 2)

,

(8, 2)

,

(4, 6)

.

Dev's work, line by line

Line	What Dev wrote
1	Outer rectangle: $(9 - 1) \times (7 - 1) = 8 \times 6 = 48$
2	Triangle: base $8 - 2 = 6$ , height $6 - 2 = 4$
3	Triangle area: $6 \times 4 = 24$
4	Area between: $48 - 24 = 24$ square units

Worked example

Find the mistake in Dev's work

Dev's four lines are in the table above. The rectangle has corners $(1, 1)$ , $(9, 1)$ , $(9, 7)$ , $(1, 7)$ and the triangle has corners $(2, 2)$ , $(8, 2)$ , $(4, 6)$ . Dev's answer is $24$ square units. Which line has a mistake? Explain what went wrong, and find the correct area between the two figures.

Check the plan. The rectangle spans x from 1 to 9 and y from 1 to 7. The triangle's corners have x-coordinates 2, 8 and 4 and y-coordinates 2, 2 and 6, all inside those ranges. All three corners are inside the rectangle, so the whole triangle is too. Subtracting in line 4 is the right plan.
Check line 1. The rectangle's side from (1, 1) to (9, 1) is horizontal and its side from (9, 1) to (9, 7) is vertical, so they meet at a right angle and their lengths multiply to give the area. Line 1 is correct.
$(9 - 1) \times (7 - 1) = 8 \times 6 = 48$
Check line 2. The base from (2, 2) to (8, 2) is horizontal. The height goes from the top corner (4, 6) straight down to (4, 2) on the base. That segment is vertical, so it is perpendicular to the base. Line 2 is correct.
$8 - 2 = 6 \quad \text{and} \quad 6 - 2 = 4$
Check line 3. This is the mistake. Multiplying base by height, 6 times 4, gives the area of the whole 6-by-4 rectangle that fits around the triangle. A triangle's area is half of its base times its perpendicular height, so Dev left out the one half.
$\frac{1}{2} \times 6 \times 4 = \frac{1}{2} \times 24 = 12$
Line 4 subtracts, which is right, but it uses the wrong triangle area. Recalculate it with 12.
$48 - 12 = 36$
Use the add-back check. The region between and the triangle together must make up the whole rectangle.
$36 + 12 = 48$

Answer

The mistake is in line 3: Dev left out the $\frac{1}{2}$ in the triangle's area. The triangle's area is $12$ square units, so the area between the two figures is $36$ square units, not $24$ .

Takeaway. When one figure lies completely inside another, the area between them is the outer area minus the inner area. Find each area first, using lengths that match the figure and heights that are perpendicular to their bases. Then subtract areas, never side lengths. To check a solution, test the plan, the measurements and the formulas one line at a time.