Area Between Two Shapes and Error Analysis | Grade 7 Math | LevelKite

You already find the area of a composite figure, a figure made of simpler shapes, by adding the areas of its parts. This lesson turns that idea around: sometimes a part is missing.

Picture a rectangular lawn with a rectangular flower bed dug into it. The whole lawn rectangle is the outer shape. The flower bed is the inner shape, and it lies completely inside the outer shape. The grass that is left covers the region between the two shapes: the flat space inside the outer shape but outside the inner shape. The area between the two shapes is the area of that region.

You find that area by subtracting, and here is why. Because the inner shape lies completely inside the outer shape, the outer shape is made of exactly two parts that do not overlap: the inner shape and the region between. So, when you measure every area in the same square unit, the outer area equals the inner area plus the area between. Take the inner area away from the outer area, and the area between is what is left.

The figure below shows the lawn and the flower bed on a grid. In every figure in this lesson, the $x$ -axis runs across and the $y$ -axis runs up. Here each grid step is $1$ meter, so each grid square covers $1$ square meter.

The lawn is

12

m by

8

m and the flower bed inside it is

5

m by

3

m, with

1

grid unit for

1

m.

Worked example

Grass around a flower bed

A rectangular lawn is $12$ m long and $8$ m wide. A rectangular flower bed that is $5$ m long and $3$ m wide lies completely inside it. In the figure, the lawn reaches from $x = 1$ to $x = 13$ and from $y = 1$ to $y = 9$ . The flower bed reaches from $x = 4$ to $x = 9$ and from $y = 3$ to $y = 6$ . What is the area of the grass between the two rectangles?

Name the two shapes. The lawn is the outer shape and the flower bed is the inner shape. All four lengths are positive and measured in meters, so both areas will be in square meters.
Find the outer area. Area counts the unit squares that cover a shape. The lawn holds 8 rows of 12 squares, each 1 square meter, so with both sides in meters you multiply the length by the width.
$\text{outer area} = 12 \text{ m} \times 8 \text{ m} = 96 \text{ m}^2$
Find the inner area the same way. Use only the flower bed's own length and width, 5 m and 3 m.
$\text{inner area} = 5 \text{ m} \times 3 \text{ m} = 15 \text{ m}^2$
Subtract. The lawn is made of the flower bed plus the grass, with no overlap, so 96 equals 15 plus the grass area. Taking 15 away from both sides keeps them equal and leaves the grass area by itself.
$\text{area between} = 96 \text{ m}^2 - 15 \text{ m}^2 = 81 \text{ m}^2$
Check the size. The flower bed is inside the lawn and does not fill it, so the grass area must be more than 0 and less than the lawn's 96 square meters. The value 81 passes.
$0 < 81 < 96$

Answer

The area of the grass between the two rectangles is $81$ square meters.

Key concept

Area between two shapes

When an inner shape lies completely inside an outer shape, the outer shape is made of two parts that do not overlap: the inner shape and the region between. So the area between equals the outer area minus the inner area.

Use this rule under these conditions:

Every length is positive, and all lengths are in the same unit, such as meters.
Each area is in the matching square unit, such as square meters.
In a triangle, the base is one side and the height is the distance from that side to the opposite corner. Measure every height perpendicular to its base, which means at a right angle to it. A side that slants is not the height.

Find each area from that shape's own given measurements. Then subtract areas, never lengths.

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

The triangle has base

12

cm and perpendicular height

8

cm, and the square inside it has sides of

3

cm, with

1

grid unit for

1

cm.

Worked example

A square inside a triangle

The figure shows a square completely inside a triangle, with $1$ grid unit for $1$ cm. The triangle's base runs from $(1, 1)$ to $(13, 1)$ , so it is $12$ cm long. The top corner is at $(7, 9)$ , so the height, measured at a right angle to the base, is $8$ cm. Each slanted side is $10$ cm. The square reaches from $x = 5$ to $x = 8$ and from $y = 2$ to $y = 5$ , so each side is $3$ cm. Find the area between the triangle and the square.

Sort the given lengths. The triangle is the outer shape and the square is the inner shape. A triangle's area uses its base and the height perpendicular to that base: 12 cm and 8 cm. The 10 cm sides slant, so neither one is the height. All lengths are positive and in centimeters.
Find the outer area. A triangle covers half the area of a rectangle with the same base and the same perpendicular height, so take half of 12 times 8.
$\text{outer area} = \tfrac{1}{2} \times 12 \text{ cm} \times 8 \text{ cm} = 6 \text{ cm} \times 8 \text{ cm} = 48 \text{ cm}^2$
Find the inner area. A square is a rectangle with equal sides, so multiply its side length by itself.
$\text{inner area} = 3 \text{ cm} \times 3 \text{ cm} = 9 \text{ cm}^2$
Subtract the inner area from the outer area. The triangle is made of the square plus the region between, with no overlap, so removing the square's area leaves the area between.
$\text{area between} = 48 \text{ cm}^2 - 9 \text{ cm}^2 = 39 \text{ cm}^2$
Check the size. The square is inside the triangle and does not fill it, and 39 is more than 0 and less than 48.
$0 < 39 < 48$

Answer

The area between the triangle and the square is $39$ square centimeters.

Now switch roles and check someone else's work. Error analysis means reading a finished solution, finding a mistake, explaining why it is a mistake, and fixing it. You already did that for the student who subtracted lengths.

Judge the work by the measurements the problem gives, not by how the drawing looks. A drawing can make a slanted side look like a height. The given numbers and their names, such as base, height and side, settle the question.

Go through the solution one line at a time and ask what each number is supposed to measure. The table lists questions to ask when the inner shape lies completely inside the outer shape without filling it.

Questions for a solution where the inner shape lies completely inside the outer shape without filling it

Question to ask	What a wrong solution does
Does each area use only that shape's own given lengths?	Uses a length from the other shape
Is each height the length given as perpendicular to its base?	Uses a slanted side as the height
Is the right area rule used for each shape?	Leaves the half out of a triangle's area
Does it subtract areas, outer minus inner?	Adds the areas, or subtracts lengths first
Are all lengths positive and in one unit, with each area in square units?	Writes an area in cm instead of square cm
Is the answer more than $0$ and less than the outer area?	Gives an answer bigger than the whole outer shape

The plate is

10

cm by

8

cm, and the triangular hole inside it has base

6

cm and perpendicular height

4

cm, with

1

grid unit for

1

cm.

Worked example

Find and fix the error

A rectangular metal plate is $10$ cm long and $8$ cm wide. A triangular hole lies completely inside it. The hole's base is $6$ cm, its height perpendicular to that base is $4$ cm, and each slanted side is $5$ cm. The figure shows this with $1$ grid unit for $1$ cm. A student finds the area of metal left: plate $10 \times 8 = 80$ ; hole $\frac{1}{2} \times 6 \times 5 = 15$ ; metal left $80 - 15 = 65$ square cm. Find and fix the error.

List the given lengths with their names before you read the work. Plate: length 10 cm, width 8 cm. Hole: base 6 cm, perpendicular height 4 cm, slanted sides 5 cm. In the figure, the hole's base runs from (3, 3) to (9, 3) and its top corner is at (6, 7), which is 4 units straight above the base. All lengths are positive and in centimeters.
Check the plate line. It multiplies the rectangle's own length and width, both in centimeters, so the outer area is correct.
$\text{outer area} = 10 \text{ cm} \times 8 \text{ cm} = 80 \text{ cm}^2$
Check the hole line. It uses the base, 6, and then 5. But 5 cm is a slanted side. A triangle's area needs the height perpendicular to the base, and the problem gives that height as 4 cm. This is the error. Because 5 is more than 4, the student's hole area is too large.
Fix the hole line by using the perpendicular height, 4 cm. Half of 6 is 3, and 3 times 4 is 12.
$\text{inner area} = \tfrac{1}{2} \times 6 \text{ cm} \times 4 \text{ cm} = 3 \text{ cm} \times 4 \text{ cm} = 12 \text{ cm}^2$
Redo the last line. Subtracting the inner area from the outer area was the right operation, because the hole lies completely inside the plate. Only the number that went into it was wrong.
$\text{area between} = 80 \text{ cm}^2 - 12 \text{ cm}^2 = 68 \text{ cm}^2$
Check the size. The hole is inside the plate and does not fill it, and 68 is more than 0 and less than 80. The student's 65 passes this test too, so the size check alone would have missed the error. Checking each number against the given lengths is what caught it.
$0 < 68 < 80$

Answer

The student used a slanted side, $5$ cm, as the triangle's height. With the perpendicular height, $4$ cm, the hole's area is $12$ square centimeters, so the area of metal left is $68$ square centimeters, not $65$ .

Here is the idea to keep. When an inner shape lies completely inside an outer shape, the area between them is the outer area minus the inner area. Find each area from that shape's own given lengths, with every length positive and in one unit and every height perpendicular to its base. Then subtract the areas and give the result in square units.

When you check a solution, yours or someone else's, test each line against the given measurements, not the look of the drawing. If the inner shape does not fill the outer shape, the area between must also be more than $0$ and less than the outer area.