Chapter 6 · Solid-Figure Problems · Lesson 6.2

Track Surface Area When Solids Join or Split

连接与切割时的表面积变化

For the joining and cutting actions here, surface area changes where faces are hidden or newly exposed. Learn to keep a precise face ledger when blocks are glued, separated, or cut.

Join → hide two faces. Cut → reveal two faces.

Grade 5 enrichment35–50 minutes10 interactive missionsSelf-contained

Mission 1

Keep a surface-area change ledger

Do not recount every face from the beginning. Start with known surface areas and record only what the joining or cutting action changes.

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Separate solids

Add the surface areas of all separate solids.

starting total = SA₁ + SA₂ + ···

Join solids

Every contact area belonged to two exposed faces before joining.

joined SA = separate total − 2 × contact area

Make a full cut

The cut exposes one new face on each of the two new pieces.

new SA = old SA + 2 × cut-face area

Important: SA means surface area. After cutting, we count the total surface area of all separated pieces, including both sides of each cut; assume no material is lost to the cut. “Two faces” does not always mean two whole faces of a cuboid. The hidden or new region may be only part of a face. Always calculate its actual area.

Choose the change

One face-to-face contact hides how many exposed regions?

One complete straight cut creates how many new faces?

When two solids join, do we add or subtract twice the contact area?

When one solid is split by a full cut, do we add or subtract twice the cut area?

Mission 2

One contact hides two equal regions

A 4×3×2 cuboid and a 2×3×2 cuboid are joined along a 3×2 rectangle.

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Join the two cuboidsinteractive

The highlighted rectangle has area 3×2=6.

First cuboid: 2(4×3+4×2+3×2)=52

Second cuboid: 2(2×3+2×2+3×2)=32

Two hidden contact regions: 2×6=12

Joined surface area: 52+32−12=72

The contact is counted once on each separate solid. That is why the contact area is subtracted twice.

Complete the contact ledger

Dimensions are in units; give all areas in square units.

Contact area

Total hidden area

Separate surface-area total

Joined surface area

Mission 3

Track every contact in a four-cube solid

Four cubes have edge lengths 5, 3, 2, and 1 centimetres. Use the placement and six contact areas specified in the table below; these are not determined by the edge lengths alone.

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Assembled solidspecified contact model

Start with all four separate cubes

6(5²+3²+2²+1²)=6(25+9+4+1)=234

5-cm and 3-cm cubes3²=92×9=18

5-cm and 2-cm cubes2²=42×4=8

3-cm and 2-cm cubes2²=42×4=8

1-cm cube touches three cubes1+1+1=32×3=6

234−(18+8+8+6)=194 cm²

The same subtraction can be grouped as 3²×2+2²×4+1²×6.

Rebuild the calculation

Surface area of all four separate cubes

Total contact area, counted once per contact

Total area hidden by joining

Surface area of the joined solid

Mission 4

A full cut creates two new faces

Move the slider. Each cut plane crosses the complete 3×4 cross-section, so every cut adds two faces of area 12.

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Cut a 12×3×4 cuboid0 cuts

0 cuts

1pieces

0new faces

12area of each face

0surface-area increase

pieces = cuts + 1 new faces = 2 × cuts surface-area increase = 2 × cuts × cut-face area

These formulas apply to separate, parallel full cuts through the same cuboid, with no waste. A partial notch can create a different set of new faces.

Suppose four full cuts are made

Answer for four cuts, regardless of the slider. Give areas in square units.

Number of pieces

Number of new faces

Area of each new face

Surface-area increase

Mission 5

Work backward from a surface-area increase

A long wooden cuboid is cut into six equal lengths by five parallel cuts perpendicular to its length, without losing material. Each piece is 2 m long, and the total surface area increases by 120 m².

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Six equal piecesoriginal practice

Six pieces require five cuts. Each cut produces two new cross-sectional faces.

Count cuts: 6−1=5.

Count new faces: 5×2=10.

Find one cross-section: 120÷10=12 m².

Find original length: 6×2=12 m.

Find volume: 12×12=144 m³.

Complete the backward chain

Give area in m², length in m, and volume in m³.

Cuts

New faces

Cross-sectional area

Original length

Original volume

Mission 6

Three cut directions reveal the three face areas

Three identical cuboids are each cut completely through once. Each cut is parallel to a different pair of the cuboid’s opposite faces; no material is lost. Their surface areas increase by 40, 48, and 60 cm².

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Pair-product puzzlepractice problem

2040÷2

2448÷2

3060÷2

Try a common factor of 20 and 24 as one edge. With 4, the other edges are 20÷4=5 and 24÷4=6. Check the third product: 5×6=30. Thus all three face areas match: 20, 24, 30:

4×5=20, 4×6=24, 5×6=30 dimensions = 4 cm, 5 cm, 6 cm surface area = 2(20+24+30)=148 cm² volume = 4×5×6=120 cm³

Recover the cuboid

Give areas in cm² and volume in cm³; list face areas in the same order as the three increases.

First face area

Second face area

Third face area

Surface area

Volume

Which dimensions fit all three products?

Mission 7

Work backward from a removed end slice

A cuboid loses an end slice of volume 32 cm³. The remaining solid is a cube of edge 4 cm.

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Original cuboid and removed slicepractice problem

The remaining cube gives the cross-section: 4×4=16 cm².

Removed length: 32÷16=2 cm.

Original length: 4+2=6 cm.

Original dimensions: 4×4×6.

Original surface area: 2(16+24+24)=128 cm².

Why the cross-section is 4×4: the remaining part is a cube, and the removed piece is an end slice of the original cuboid.

Recover the original solid

Give lengths in cm and areas in cm².

Cross-sectional area

Removed length

Original length

Original surface area

Mission 8

Same blocks, same volume, different surface areas

Three identical 4×3×2 cuboids form one larger cuboid. All dimensions are in centimetres. Keep the blocks in the same orientation and join whole matching faces in a straight line; compare the three arrangements A, B, and C.

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Arrangement A12×3×2

Surface area: 132 cm² · Volume: 72 cm³

Arrangement	Dimensions	Surface area	Volume
A	12×3×2	132	72
B	4×9×2	124	72
C	4×3×6	108	72

All three arrangements use the same three blocks, so volume stays constant. Among these three arrangements, A has the greatest surface area and C the least. Compare their calculated areas, not their appearance.

Compare the arrangements

Which arrangement has the greatest surface area?

Greatest surface area

Do the three volumes differ?

Smallest surface area

Mission 9

Surface-area change workshop

Part A considers three separate changes to the same original cuboid: increase only its length by 5 cm (volume rises by 150 cm³), only its width by 4 cm (volume rises by 160 cm³), or only its height by 3 cm (volume rises by 144 cm³). Then solve eight mixed problems. All twelve answers are needed for this checkpoint.

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Strategy reminder: a volume increase is a new slice

When only one dimension grows, the added volume equals the unchanged cross-sectional area times the added length. Divide the volume increase by the length increase to recover that face area.

Then account for both copies of each face

A cuboid has three pairs of equal opposite faces. Once you know one area from each pair, add those three areas and double the sum. You do not need the individual edge lengths.

Part A · Calculate the face areas

150÷5

160÷4

144÷3

Original surface area

1. Two cubes of edge 3 cm are joined on one full face. Find the surface area.

2. A cuboid is cut into 4 pieces by 3 parallel full cuts. Each cut face has area 15 cm². How much does surface area increase?

3. Four joined cubes have edge lengths 5, 3, 2, and 1 cm as in Mission 3. Find the surface area.

4. The six-piece wooden cuboid in Mission 5 has what volume?

5. Using the three face-parallel cuts described in Mission 6, the increases are 40, 48, and 60 cm². Find the original surface area.

6. A 4-cm cube remains after an end slice of volume 32 cm³ is removed. Find the original surface area.

7. Three 4×3×2 blocks form a cuboid. Among the three straight-line arrangements A, B, and C in Mission 8, what is the greatest surface area (cm²)?

8. Dimension increases reveal face areas 30, 40, and 48 cm². Find the original surface area.

Mission 10

Exit ticket

Answer all five questions correctly to earn the Surface-Area Change Engineer certificate.

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1. A contact rectangle is 4×3. By how much does joining reduce the separate surface-area total?

2. A cuboid is cut into 7 pieces by parallel full cuts. How many new faces are created?

3. What is the surface area (cm²) of the four-cube arrangement with the six contacts listed in Mission 3?

4. Using the three face-parallel cuts in Mission 6, if the increases are 40, 48, and 60 cm², what is the original cuboid’s surface area?

5. A cuboid’s three face areas are 30, 40, and 48. What is its surface area?

Optional reflection — not automatically graded

Teaching notes

The contact machine, dynamic cut model, exploded diagrams, pair-product puzzle, original-independent explanations, diagnostic questions, and exit ticket are additional instructional scaffolds. Diagrams are redrawn and are not necessarily to scale.