Perimeter and Area · Lesson 4 of 7
Same Area, Different Perimeter
“Explore how rearranging pieces preserves area while changing their outside boundary.”
• Compare tangram piece areas using a common small triangle. • Explain area conservation when pieces are rearranged without gaps or overlaps. • Compare whole-number rectangles of fixed area. • Find perimeter extremes for nine connected unit squares. • Predict perimeter changes when a square is attached along one, two or three edges.
The same pieces can cover different outlines
When you rearrange a paper puzzle, you move its material rather than creating new material. If every piece is used once without gaps or overlaps, the total area stays the same. The outside boundary can change because different piece edges become exposed. A tangram makes this difference visible.
A puzzle made from seven pieces that can be rearranged to form different figures.
Take small triangle C as one area unit. Small triangle E covers the same area, so it is also one unit. Together C and E can cover square D exactly, showing that D has area two units. The medium triangle F and parallelogram G can each be matched to the same two-unit area, even though their shapes differ.
The large triangles A and B each cover the area of four small triangles. One way to see the comparisons is to place matching pieces over one another or cut paper copies into smaller triangles. The small triangle here is an area unit for this puzzle, not a centimetre measurement; its physical area depends on how large the tangram is.
| Piece | Area in units of C | Relationship |
|---|---|---|
| C, E | 1 each | Equal small triangles |
| D, F, G | 2 each | Each twice C |
| A, B | 4 each | Each twice G and four times C |
Problem
Find the area of the square formed by all seven pieces, in units of C.
- 1.Add all piece areas: A 4, B 4, C 1, D 2, E 1, F 2 and G 2.
- 2.Total = 4 + 4 + 1 + 2 + 1 + 2 + 2 = 16 units of C.
- 3.If the same seven pieces make a rectangle without gaps or overlaps, that rectangle also has area 16 units of C.
- 4.Compare its boundary with the square’s by tracing or measuring; equal area alone does not force equal perimeter.
Cut paper copies of the seven labelled pieces. Verify C = E, D = F = G and A = B by matching or rearranging pieces. Build a square, then a rectangle. Explain why the area stays 16 units of C, and measure the two perimeters rather than assuming they match.
Fixed area and whole-number rectangle sides
We can investigate the same idea without cutting paper. Keep a rectangle’s area fixed and change its length and width. If both dimensions must be whole numbers, list the factor pairs of that area. Every pair gives the same product but can give a different sum and therefore a different perimeter.
Problem
Find rectangle dimensions with area 24 square units and compare their perimeters.
- 1.A pair of positive whole-number dimensions must multiply to 24. Ignoring rotations, the pairs are 1 and 24, 2 and 12, 3 and 8, and 4 and 6.
- 2.Their perimeters are 2 × 25 = 50, 2 × 14 = 28, 2 × 11 = 22, and 2 × 10 = 20 units.
- 3.The long thin 1 by 24 rectangle has greatest perimeter 50. The more balanced 4 by 6 rectangle has least perimeter 20 among these whole-number choices.
- 4.All four rectangles still cover 24 unit squares.
| Dimensions | Area | Perimeter |
|---|---|---|
| 1 × 24 | 24 square units | 50 units |
| 2 × 12 | 24 square units | 28 units |
| 3 × 8 | 24 square units | 22 units |
| 4 × 6 | 24 square units | 20 units |
Problem
Repeat the investigation for area 32 cm², using positive whole-number side lengths.
- 1.The distinct factor pairs are 1 × 32, 2 × 16 and 4 × 8.
- 2.Their perimeters are 66 cm, 36 cm and 24 cm.
- 3.The 1 cm by 32 cm rectangle has greatest perimeter. The 4 cm by 8 cm rectangle has least perimeter among these whole-number choices.
- 4.The pattern suggests that the closest factor pair gives the smaller boundary, while the most unequal pair gives the larger one. Listing all factor pairs checks the claim for this area.
The whole-number condition matters because it specifies which rectangles we are comparing. Turning 4 by 8 into 8 by 4 gives a rotation, not a new area-and-perimeter result. When investigating another whole-number area, write all its factor pairs before choosing a greatest or least perimeter.
Problem
Make rectangle A of area 18 square units and rectangle B of area 20 square units, with A having the larger perimeter.
- 1.Choose A as 2 units by 9 units. Its area is 18 and its perimeter is 2 × (2 + 9) = 22 units.
- 2.Choose B as 4 units by 5 units. Its area is 20 and its perimeter is 2 × (4 + 5) = 18 units.
- 3.Thus A has less area but more perimeter. A long thin shape can need more boundary than a compact larger shape.
Equal perimeter does not force equal area either. A 1 by 5 rectangle and a 2 by 4 rectangle both have perimeter 12 units. Their areas are 5 and 8 square units. Area comes from the product of dimensions, whereas rectangle perimeter comes from twice their sum.
Nine unit squares: exposed edges decide perimeter
Now keep nine identical unit squares and join them edge to edge as one connected region without holes. All arrangements have area nine square units. To find perimeter, count exposed unit edges. A shared edge belongs inside the figure and contributes nothing to its boundary.
A 3 by 3 square has perimeter 12 units. A row of nine squares makes a 9 by 1 rectangle with perimeter 20 units. In the row, each added square attaches along one edge, increasing the boundary by two units. A connected group of nine squares must have at least eight shared edges, so it cannot exceed the row’s perimeter: 4 × 9 − 2 × 8 = 20.
The compact 3 by 3 arrangement also gives the minimum, 12. To see why, suppose an arrangement uses r occupied rows and c occupied columns. Each row needs at least a left and right exposed edge, and each column needs a top and bottom exposed edge, so perimeter is at least 2(r + c). If r + c were at most 5, the largest possible row-column grid would have only 2 × 3 = 6 positions. It could not contain nine squares. Thus at least 12 exposed edges are needed, and the 3 by 3 square achieves them.
Problem
Arrange nine unit squares as one connected shape with perimeter 18 units.
- 1.Make a row of seven squares. Attach two more immediately above the first two positions.
- 2.The long row has six shared edges. The upper pair has one shared edge with each other and two contacts with the lower row: total shared edges = 9.
- 3.The nine separate squares have 36 edges. Remove two copies for each shared edge: perimeter = 36 − 2 × 9 = 18 units.
- 4.The area remains nine square units and there are no holes.
There is more than one nine-square shape with perimeter 20: a straight row and an L-shaped chain both have only eight shared edges. A 3 by 3 square is the only minimum-perimeter shape apart from moving or turning it. A perimeter of 18 also allows different arrangements; trace exposed edges to check each construction.
Adding area can even shorten the boundary
An added square always contributes one extra square unit of area if it does not overlap. Its effect on perimeter depends on how many edges touch the existing figure. Each contact removes an old boundary edge and hides one edge of the new square. This can outweigh the new exposed edges.
Starting with perimeter 24 units, one, two or three full edge contacts would give new perimeters 26, 24 or 22 units. You can decide from the contacts without recounting the whole boundary.
With one contact, the new square exposes three edges but removes one old edge: the net change is +2. With two contacts, it exposes two and removes two: no change. With three contacts, it exposes one and removes three: the boundary decreases by 2. Filling an inward notch can therefore increase area while shortening the perimeter.
More area does not automatically mean more perimeter. Check the outer edges. Likewise, rearranging the same pieces preserves area only if you use every piece once without gaps or overlaps.
Half an area and separate boundaries
Dividing a shape keeps track of both what is covered and what becomes exposed. Half the area does not require half the perimeter. When pieces are separated, their new cut edges become parts of their individual boundaries. A smaller region placed inside another shape also needs its own dimensions.
Problem
A rectangle is 12 cm by 8 cm. Draw an inner rectangle of half its area without touching its boundary.
- 1.The outer area is 12 × 8 = 96 cm², so half is 48 cm².
- 2.Choose inner dimensions 8 cm by 6 cm, since 8 × 6 = 48 cm². Both dimensions are smaller than the corresponding outer dimensions.
- 3.Centre it to leave 2 cm at each end of the 12 cm direction and 1 cm at each end of the 8 cm direction.
- 4.A 12 cm by 4 cm choice would have the right area but would touch the outer boundary, so it does not meet the condition.
Problem
A square of side 8 cm is divided into two equal rectangles. Compare their total area and the sum of their separate perimeters with the original.
- 1.Original area = 8 × 8 = 64 cm²; perimeter = 4 × 8 = 32 cm.
- 2.Each rectangle is 8 cm by 4 cm, with area 32 cm² and perimeter 2 × (8 + 4) = 24 cm.
- 3.Areas add to 64 cm², but separate perimeters add to 48 cm, which is 1½ times 32 cm.
- 4.The new cut has length 8 cm and is counted on both separated pieces, adding 16 cm to the total boundary.
Quiz
If C has area 1 small-triangle unit, what is the tangram’s total area?
Which whole-number rectangle of area 24 has least perimeter?
What is the greatest perimeter of nine connected unit squares with no holes?
A new unit square shares three edges. How does perimeter change?
Which statement is always true when all puzzle pieces are rearranged without gaps or overlaps?
Which pair shows equal perimeter but different area?
Practice Problems
- Use paper pieces to explain why tangram D, F and G have the same area.
- List every positive whole-number rectangle of area 36 square units and find the greatest and least perimeters in that list.
- Construct two different nine-square shapes with perimeter 18 units. Verify their exposed edges.
- Draw an example where adding one square leaves perimeter unchanged and another where it decreases.
- Find two rectangles with perimeter 16 units and different areas.
- A 10 cm square is divided into equal rectangles by one fold. Find each rectangle’s area and the sum of the separate perimeters.
- Inside a 14 cm by 10 cm rectangle, choose an inner rectangle of half the area that touches none of the outer sides.
Key Takeaways
• Rearranging all pieces without gaps or overlaps preserves area but may change perimeter. • Tangram area comparisons use the small triangle as a common unit. • Fixed area uses factor pairs; compare their sums to compare perimeters. • Nine connected unit squares have minimum perimeter 12 and maximum 20 unit lengths. • One, two or three contacts for an added unit square change perimeter by +2, 0 or −2. • Equal area and equal perimeter are different conditions; neither implies the other.