Twenty-Four Tiles, Thirty Panels: Area and Perimeter Under a Constraint
Mrs. Bianchi's class is planning rectangular garden beds for the school courtyard. Each bed must be covered with exactly 24 square tiles that are 1 foot on each side, and each bed must be enclosed by 1-foot fence panels. The class has 30 panels in the shed and cannot buy more. Groups propose these beds: 1 by 24, 2 by 12, 3 by 8, and 4 by 6. One group claims the 1-by-24 bed is best 'because it holds the most.' Another group says all four beds use the same amount of fence 'because they all use 24 tiles.' A third group finds the perimeter of the 4-by-6 bed by adding 4 + 6 and gets 10 panels. Mrs. Bianchi must help the class decide which designs are actually buildable with the panels on hand and which reasoning needs reteaching before the beds are marked out in the courtyard.
1. Compute the area and perimeter of all four proposed beds and determine which designs can actually be built.
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Every design uses 24 tiles, so all four have an area of 24 square feet. The perimeter is 2 × (length + width).
| Bed | Area (tiles) | Perimeter (panels) | Buildable with 30 panels? |
|---|---|---|---|
| 1 × 24 | 24 | 2 × (1 + 24) = 50 | No - 20 panels short |
| 2 × 12 | 24 | 2 × (2 + 12) = 28 | Yes, with 2 left over |
| 3 × 8 | 24 | 2 × (3 + 8) = 22 | Yes, with 8 left over |
| 4 × 6 | 24 | 2 × (4 + 6) = 20 | Yes, with 10 left over |
Three designs are buildable. The 1-by-24 bed is not, and the class should notice why: as the rectangle gets longer and thinner, more of the 24 tiles end up on the outside edge, and each edge tile needs fence. The square-like 4-by-6 bed uses the fewest panels - 20 - while covering exactly the same ground.
2. Respond to the two claims: 'the 1-by-24 bed is best because it holds the most,' and 'all four beds use the same fence because they all use 24 tiles.'
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'The 1-by-24 holds the most.' False. All four beds hold exactly 24 tiles - area is fixed by the problem. Holding the most and needing the most fence are being confused with each other. The 1-by-24 is in fact the worst option available: it needs 50 panels, more than the class owns, and a 24-foot strip is unlikely to fit the courtyard. Have the group count tiles in two designs side by side; the count is identical.
'They all use the same fence.' Also false, and this is the more important misconception. Equal area does not force equal perimeter. A useful way to show it: start with the compact 4-by-6 and move one tile from a short side to a long side, turning it toward 3-by-8. Area is unchanged at 24, but the two tiles that move each trade a shared edge for an exposed one, so the perimeter grows from 20 to 22. Repeat toward 2-by-12 (28) and 1-by-24 (50). The perimeter grows every time the shape gets thinner.
Both claims come from one root idea: treating area and perimeter as the same measurement because both are computed from the same two side lengths.
3. A group computes the perimeter of the 4-by-6 bed as 4 + 6 = 10 panels. Diagnose the error and describe a representation that fixes it.
Read the worked answer
The group added one length and one width - an L-shaped trip around two sides of the bed. They stopped at the corner instead of going all the way around. Ten panels would leave the bed fenced on two sides and open on the other two.
Fix it physically first. Have them walk the edge of a taped 4-by-6 rectangle, laying down 1-foot paper panels as they go and counting aloud: 4 across, 6 down, 4 back, 6 up - 20. The moment they turn the second corner they will see that the sides they have already counted have matching sides opposite them. Then ask, 'how many sides does a rectangle have?' before returning to the numbers: 4 + 6 + 4 + 6.
Then connect to the formula. Write the walk as 2 × 4 + 2 × 6 and as 2 × (4 + 6), and have the group explain why doubling the 10 is legitimate - because every rectangle has two of each side. A grid drawing with a dot marked on each unit of edge gives the same count without the walking, and can be used as the check on the day of building.