Chapter 4 · Plane Geometry

Lesson 4.6 — Equal-Area Partitions and Missing Lengths

第4讲 · 等面积分割与缺失长度

A complicated diagram often becomes manageable after its total area is divided into equal-area groups. Once each group’s area is known, the triangle and rectangle formulas can be used backward to reveal missing lengths.

Find the area share first. Let that area reveal the lengths.

Grade 5 enrichment 55–70 minutes Everything needed is on this page. Autosaves in this browser

Mission 1

Treat the whole area as a budget

Not completed

When equal-area groups cover the whole figure without gaps or overlap, the whole area is shared equally. The groups do not have to look alike.

Whole area 54

Equal groups 3

54 ÷ 3 = 18 square units per group.

For the current sliders, what is the area of each group?

square units

If one equal group is made from two smaller pieces, what is the total area of those two pieces?

What is the total area of all the groups together?

square units

Mission 2

One equal group may contain several visible pieces

Not completed

The statement A = B = C + D names three equal groups: Group A, Group B, and the combined group C + D.

Four visible pieces, but only three equal-area groups.

Whole area: 48 square units

Equal groups: A, B, and C + D

Area per group: 48 ÷ 3 = 16

Area of A

square units

Area of C + D

square units

Ignore the example’s labels C = 6 and D = 10 for this question. From A = B = C + D alone, can the exact area of C be determined?

Mission 3

Use a known area backward to recover a length

Not completed

Triangle height

h = 2A ÷ b

If area and base are known, double the area and divide by the base.

Triangle base

b = 2A ÷ h

If area and height are known, double the area and divide by the height.

Leftover segment

whole − known part

Area often reveals one segment, and subtraction reveals the neighboring segment.

A triangle has area 18 square units and base 9 units. Its height is:

units

A triangle has area 18 square units and perpendicular height 6 units. Its base is:

units

A full side is 6 units. One part is 4 units. The leftover part is:

units

Do equal areas always force two lengths to be equal?

Mission 4

Find the three equal-area groups in a 9 × 6 rectangle

Not completed

Rectangle ABCD has AB = DC = 9 cm and AD = BC = 6 cm. E lies on BC and F on DC. Its four regions fill the rectangle without gaps or overlap. S₁, S₂, S₃, and S₄ denote their areas, and the given condition is:

S₁ = S₂ = S₃ + S₄

The third equal group is a compound region: S₃ + S₄.

9 × 6 = 54

54 = S₁ + S₂ + (S₃ + S₄)

S₁ = S₂ = S₃ + S₄ = 18

The diagram is not drawn for measuring. Use only stated dimensions and proved relationships.

Area of the rectangle

cm²

Number of equal-area groups

Area of each equal group

cm²

Which expression names the third equal group?

Mission 5

Use the 18 cm² area share to recover BE and DF

Not completed

S₁ and S₂ are triangles. Their known areas turn into equations for their missing perpendicular dimensions.

Triangle S₁

S₁ = ½ × AB × BE

18 = ½ × 9 × BE

BE = 4 cm

Triangle S₂

S₂ = ½ × AD × DF

18 = ½ × 6 × DF

DF = 6 cm

BE = 4from area S₁

EC = 2because BC = 6

DF = 6from area S₂

FC = 3because DC = 9

EC = BC − BE

FC = DC − DF

Mission 6

Use the leftover lengths to separate S₃ and S₄

Not completed

S₃ = ½ × EC × FC

S₃ = ½ × 2 × 3 = 3 cm²

S₄ = (S₃ + S₄) − S₃

S₄ = 18 − 3 = 15 cm²

Area of S₃

cm²

Area of S₄

cm²

Why is S₄ found by 18 − S₃?

Mission 7

Three equal regions inside a right trapezoid

Not completed

Right trapezoid ABCD has parallel bases AD = 12 cm and BC = 15 cm, with perpendicular height AB = 8 cm. E lies on AB and F on BC. Triangle ADE, quadrilateral DEBF, and triangle CDF have equal areas. Find the area of triangle EBF.

Trapezoid area = ½ × (12 + 15) × 8 = 108

Each equal region = 108 ÷ 3 = 36

½ × 12 × (8 − BE) = 36, so BE = 2

½ × FC × 8 = 36, so FC = 9 and BF = 6

Area of EBF = ½ × 2 × 6 = 6

Area of each of the three equal regions

cm²

Area of triangle EBF

cm²

Mission 8

Follow an area chain through a square

Not completed

Square ABCD has side 12 cm. E lies on DC with DE twice EC. The line through A and E meets the extension of BC beyond C at F. Find the area of triangle DEF and then the length CF.

For triangle ADF, choose AD as the 12 cm base. Since F lies on the line BC, parallel to AD, its perpendicular height is also 12 cm. For triangle DEF, DE is the base and CF is the perpendicular height.

DE + EC = 12 and DE = 2EC

DE = 8, EC = 4

Area ADF = ½ × 12 × 12 = 72

Area ADE = ½ × 12 × 8 = 48

Area DEF = 72 − 48 = 24

24 = ½ × 8 × CF, so CF = 6

Area of triangle ADF

cm²

Area of triangle DEF

cm²

Mission 9

Equal-area and missing-length workshop

Not completed

Solve at least six of the eight questions. All lengths below are in cm and areas in cm². Find the equal-area group’s share before doing length arithmetic.

1A figure of area 84 is split into 4 equal groups. Area per group?

2Four pieces cover a figure of area 75 cm² without gaps or overlap. Their areas satisfy A = B = C + D. Find A.

3Triangle area 27, base 9. Find the height.

4A 10 × 6 rectangle forms 3 equal groups. One group is a triangle with base 10. Find its height.

5In Mission 4’s 9 × 6 rectangle, find BE.

6In that rectangle, find S₄.

7In Mission 7’s right trapezoid, area of triangle EBF?

8In Mission 8’s square, find CF.

A final area chain outside a rectangle

Exercise 12

Rectangle ABCD has AB = 8 cm and BC = 6 cm. F lies on AD with AF = 4 cm. E lies beyond D on the line CD, and E, F, B lie in that order on one straight line. Find the area of shaded triangle AEF.

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

Area of ABF (cm²)

Area of EAB (cm²)

Area of AEF (cm²)

Hint — choose a first step

AB is a useful common base. E lies on the same horizontal line as D and C. Compare triangles EAB and FAB.

Worked solution — compare after trying

Area ABF = ½ × 8 × 4 = 16.
Triangle EAB has base AB = 8 and height 6, because E lies on the lower base line. Its area is 24.
F is on EB, so AF partitions triangle EAB into ABF and AEF. The shaded area is 24 − 16 = 8 cm².

Worked workshop solutions — open after attempting

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

Share the area: 84 ÷ 4 = 21.
There are three groups, A, B, and C + D. Each has area 75 ÷ 3 = 25.
Double the area and divide by the base: 27 × 2 ÷ 9 = 6.
Each group has area 60 ÷ 3 = 20. The triangle height is 20 × 2 ÷ 10 = 4.
S₁ = 18 and base AB = 9, so BE = 18 × 2 ÷ 9 = 4.
EC = 2 and FC = 3, so S₃ = 3. Remove it from the 18-unit compound group: S₄ = 15.
Each group has area 36. This gives BE = 2 and BF = 6; EBF = 2 × 6 ÷ 2 = 6.
DE = 8. Area DEF = 72 − 48 = 24. Thus CF = 24 × 2 ÷ 8 = 6.

Mission 10

Exit ticket

Not completed

Try all five questions without hints first. If you need help, review the relevant mission and then retry.

Equal-area groups cover an entire figure without gaps or overlap, and its total area is known. How can you find each group’s area?

In Mission 4’s 9 × 6 rectangle, each equal group has area:

cm²

In the same rectangle, BE equals:

In Mission 7’s right trapezoid, triangle EBF has area:

cm²

In Mission 8’s square, CF equals:

Optional reflection — not automatically graded: How did a known area help you find a missing length?

Equal-Area Partition Architect

You have completed this lesson.

You can divide a whole into equal-area groups, treat compound regions as one group, and use area formulas backward to recover missing lengths.

Teaching notes

The original rectangle is 9 cm by 6 cm and satisfies S₁ = S₂ = S₃ + S₄.Its area is 54 cm², so each of the three equal groups has area 18 cm².The original then obtains BE = 4 cm, DF = 6 cm, EC = 2 cm, FC = 3 cm, S₃ = 3 cm², and S₄ = 15 cm².

The final external-triangle investigation covers Exercise 12 with a fresh area chain.The area-budget model, compound-group introduction, backward-formula review, feedback, workshop, and exit ticket are added instructional scaffolds.Diagrams are explanatory and are not scale drawings.