Chapter 4 · Plane Geometry

Lesson 4.4 — Overlap and Cancel Common Areas

第4讲 · 重叠图形与公共部分消去

When two equal-area figures overlap, the shared region can hide a powerful shortcut. Remove the same overlap from both wholes, and the two exposed remainders must have equal area.

Equal wholes − the same overlap = equal remainders.

Grade 5 enrichment 45–60 minutes Layered SVG diagrams Autosaves in this browser

Mission 1

See the three layers in an overlap problem

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Before calculating, name what you see. Each complete figure is a whole. The part inside both figures is the common overlap. What remains after the overlap is removed is an exposed remainder.

Remainder A Overlap C Remainder B

Whole A = A + C and Whole B = B + C

1. Which region belongs to both complete rectangles?

2. What do we call A and B after the common overlap is removed?

3. If the two wholes have equal area, what must be true after the same C is removed?

Mission 2

Treat area like a balanced equation

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The cancellation is not a picture trick. It is the subtraction property of equality written with areas.

A + C = 48

B + C = 48

Area of each whole48

Area of common overlap C17

Whole A

31 + 17

Whole B

31 + 17

Subtract 17 from both equal wholes: A = B = 31 square units.

Start with equal wholes.
A + C = B + C

Remove the same C from both sides.
A + C − C = B + C − C

The remainders are equal.
A = B

1. For the slider values now shown, what is each remainder?

square units

2. Which operation protects the equality?

3. The conclusion is:

Mission 3

Know when cancellation is valid

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The word same matters twice: the complete figures must have equal area, and the removed region must be the same overlap—or at least have a proved equal area on both sides.

Case A · Congruent triangles share one overlap

The complete triangles are congruent (exactly the same shape and size, possibly moved or turned), and the center region belongs to both.

Case B · Two pieces merely look alike

The two separate pieces are not the same region, and no equal-area fact is given.

Case C · The same overlap, unequal wholes

One whole has area 50 and the other has area 44. Both contain the same overlap.

Case D · Different shapes, equal whole areas

A triangle and a parallelogram each have area 36 and share the exact same overlap.

Mission 4

Build the chapter’s overlapping-triangle diagram

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Worked example 2 Two congruent right triangles overlap. Triangle ABC consists of Regions 1 and 2; triangle DEG consists of Regions 2 and 3. AB = DE = 8 dm, DF = 3 dm, and BE = 3 dm. B, E, C, and G lie on the same straight line; AB and DE are perpendicular to it.

Diagram not to scale. The two complete right triangles are congruent.

First complete triangle:Area(ABC) = Area(1) + Area(2)

Second complete triangle:Area(DEG) = Area(2) + Area(3)

The complete triangles are congruent:Area(1) + Area(2) = Area(2) + Area(3)

Remove the exact same Area(2):Area(1) = Area(3)

1. Which region is common to both complete triangles?

2. Which equation describes the two congruent wholes?

3. After cancelling Region 2, what follows?

Mission 5

Calculate the easier remainder, then transfer its area

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Use the diagram in Mission 4. Region 3 has the same area as trapezoid ABEF (Region 1). Congruence gives DE = AB = 8 dm. Since F lies between D and E and DF = 3 dm, EF = DE − DF = 5 dm. AB and EF are parallel, and BE = 3 dm is their perpendicular separation.

Region 1 is trapezoid ABEF. Its parallel sides are vertical.

Step	Calculation
Find EF	8 − 3 = 5 dm
Area of Region 1	(8 + 5) × 3 ÷ 2 = 19.5 dm²
Transfer by cancellation	Area(3) = Area(1) = 19.5 dm²

Awkward Region 3 → equal, easier Region 1 → 19.5 dm²

1. EF = 8 − 3 =

2. Area of trapezoid ABEF:

dm²

3. Therefore Area(3) is:

dm²

Mission 6

Move equal wholes—the equality survives

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Slide one rectangle of width 8 units and height 5 units horizontally. The overlap changes, but the left and right exposed strips always have the same area.

Horizontal shift3

Left remainder15

Overlap25

Right remainder15

1. For the current shift, what is each exposed remainder’s area?

square units

2. As the second rectangle moves, are the two exposed remainders always equal?

3. Why?

At shift 3, each exposed strip has area 15. The overlap may change, but it is still the same region inside both equal wholes.

Mission 7

Apply cancellation to two congruent trapezoids

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original Exercise 7 All lengths in this diagram are in cm. The upper and lower trapezoids are congruent. Their overlap is common, so the shaded exposed part of the upper trapezoid equals the exposed lower remainder.

The dashed base of the upper trapezoid is 10. Two units lie beyond the lower trapezoid, leaving 8 inside.

Cancel the common overlap.
The two exposed remainders have equal area.

Find the shorter base of the easier remainder.10 − 2 = 8 cm

Use the trapezoid formula.(10 + 8) × 5 ÷ 2 = 45 cm²

1. The upper base of the easier lower remainder is:

2. Its area is:

cm²

3. The original shaded area is:

Mission 8

Cancellation detective

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Decide what the area facts really allow you to conclude. Do not rely on how a picture looks.

1 · Equal wholes, overlap 12

Each whole has area 35.

2 · Same overlap, unequal wholes

The wholes have areas 50 and 44.

3 · Equal-area wholes, different shapes

Both wholes have area 60 and share one overlap of area 18.

4 · Congruent wholes

Why is congruence useful here?

5 · One shared region

What must be removed from both whole-area equations?

6 · Irregular remainders

Can irregular exposed regions still be equal in area?

Mission 9

Overlap-and-cancellation workshop

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Reach at least 6 out of 8. All answers are numerical or selected; no free-text response is automatically graded.

1Two equal-area figures each have area 41 and share an overlap of area 16. Find the area of each exposed remainder in square units.

2Two rectangles each have width 9 units and height 4 units. Starting exactly on top of one another, one moves 3 units horizontally with no vertical movement. Find each exposed strip’s area in square units.

3In Mission 4’s triangle diagram, EF = DE − DF = 8 − 3. Find EF in dm.

4In Missions 4–5, find Region 3’s area in dm².

5Equal wholes share an overlap. One exposed remainder has area 27. The other has area:

6In Mission 7, find the shorter base of the easier lower remainder in cm:

7In Mission 7, find the shaded area in cm².

8Wholes have areas 64 and 64. Their common overlap is 19. Find the area of each remainder.

Hints

1, 5, 8: whole − overlap. 2: exposed width × rectangle height. 3–4: use the worked example trapezoid ABEF. 6–7: 10 − 2, then the trapezoid formula.

Two adjacent squares and one shaded triangle

Exercise 11

ABCD is a square of side 5 cm. CGFE is a square of side 4 cm. B, C, and E lie in that order on one straight line. The squares lie on the same side of that line, as shown. Find the area of shaded triangle AGE.

Use the marked dimensions and relationships; do not measure the drawing.

Area of ABE (cm²)

Area of ADG (cm²)

Area of GFE (cm²)

Area of AGE (cm²)

Hint — choose a first step

The two squares together are a familiar whole. Remove triangle ABE and the two unshaded corner triangles ADG and GFE.

Worked solution — compare after trying

Total square area = 25 + 16 = 41. BE = 9, so ABE has area ½ × 9 × 5 = 22.5.
DG = 5 − 4 = 1. Area ADG = ½ × 5 × 1 = 2.5. Area GFE = ½ × 4 × 4 = 8.
Subtract each unwanted piece once: 41 − 22.5 − 2.5 − 8 = 8 cm².

Worked workshop solutions — open after attempting

Find the first step that differs from your own work, then try the problem again.

Subtract the overlap: 41 − 16 = 25.
Exposed strip width 3, height 4: area = 12.
EF = 8 − 3 = 5.
Cancel the overlap, then use the easier trapezoid: (8 + 5) × 3 ÷ 2 = 19.5.
Equal wholes lose the same overlap, so the other remainder is also 27.
The overhang is 2, so the remaining base is 10 − 2 = 8.
Replace the shaded region by the equal lower trapezoid: (10 + 8) × 5 ÷ 2 = 45.
Each whole loses 19: 64 − 19 = 45.

Mission 10

Exit ticket

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Try all five questions using the area facts, then check your answers.

1. Which statement is the cancellation principle?

2. In Missions 4–5’s overlapping-triangle example, what is the shaded Region 3 area?

dm²

3. In Mission 7’s overlapping-trapezoid example, what is the shaded area?

cm²

4. Two equal wholes each have area 52 and share an overlap of area 21. Find the area of each remainder.

square units

5. Before using cancellation to conclude that two remainders have equal area, what must you verify?

Optional reflection — not automatically graded: In one sentence, describe what you would circle first in a new overlap problem.

Overlap-Cancellation Architect

You have completed this lesson.

You can identify equal wholes, isolate a common overlap, cancel that area correctly, and replace an awkward exposed region with an equal, easier one.

Lesson 4.4 · Grade 5 Math Explorer

Teaching notes

The original divides two congruent right triangles into regions 1, 2, and 3, establishes Area(1) + Area(2) = Area(2) + Area(3), and obtains Area(1) = Area(3).It then calculates Region 1 as a trapezoid: EF = 8 − 3 = 5 and (5 + 8) × 3 ÷ 2 = 19.5 square decimetres.

Two congruent trapezoids share an overlap.The shaded remainder equals the lower exposed trapezoid, whose bases are 10 and 10 − 2 = 8 and whose height is 5, giving 45 square centimetres.

The adjacent-squares investigation adds Exercise 11 with a worked decomposition. The rectangle slider, validity cases, area-balance model, feedback, workshop, and exit ticket are added instructional scaffolds. Diagrams are explanatory and are not scale drawings.