Same Perimeter, Different Area
Students are designing a rectangular rabbit pen using 20 meters of fencing. The pen must have whole-number side lengths in meters and must enclose the greatest possible area. Student J draws a 5 m by 5 m square and says it uses 20 m of fencing and has area 25 m², so it is the best. Student K draws a 9 m by 1 m rectangle, computes perimeter 20 m and area 9 m², and says both pens use the same fencing so they are equally good. The teacher wants students to test the claim that “same perimeter means same area” and to organize possible rectangles in a table. The class has grid paper. One group claims a 6 m by 4 m pen is impossible because 6 + 4 = 10, not 20.
1. Evaluate Student J’s and Student K’s claims. Who is correct about the greatest area, and what misunderstanding does Student K show?
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Student J is correct that the 5 m by 5 m square has area 25 m² and uses 20 m of fencing. Student K is wrong to say the pens are equally good. Equal perimeter does not guarantee equal area. The 9 m by 1 m rectangle does use 20 m of fencing, but its area is only 9 m². Student K may be confusing perimeter with area or assuming that because the fencing length is the same, the enclosed space must be the same. The square encloses more area than the long, thin rectangle.
2. Make a table of all possible whole-number dimensions for a rectangle with perimeter 20 m, and identify the greatest area.
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For perimeter 20, length + width = 10. Whole-number pairs are:
| Length (m) | Width (m) | Area (m²) |
|---|---|---|
| 9 | 1 | 9 |
| 8 | 2 | 16 |
| 7 | 3 | 21 |
| 6 | 4 | 24 |
| 5 | 5 | 25 |
The greatest area is 25 m², from the 5 m by 5 m square. The table shows that as the dimensions get closer together, the area increases. It also disproves the claim that all rectangles with perimeter 20 have the same area.
3. Explain the group’s error about the 6 m by 4 m pen. What should they do to check whether it uses 20 m of fencing?
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The group added only one length and one width, getting 6 + 4 = 10. Perimeter includes two lengths and two widths, so the correct computation is 6 + 4 + 6 + 4 = 20 m, or 2 × (6 + 4) = 20 m. The 6 m by 4 m pen is possible and has area 24 m². To check any rectangle, students should add all four sides or use the perimeter formula, not add just two adjacent sides.