Perimeter, Area and the Path Problem
You teach mathematics to Class 5 in a government primary school in Udham Singh Nagar. There are 38 students. The problem on the board is: 'A rectangular field is 24 m long and 18 m wide. A path 2 m wide runs inside along the boundary. Find the area of the path.' The textbook answer is 152 m². Three students respond. S1 writes 24×18 − 20×14 = 152. S2 writes 2(24+18) = 84 and says the area is 84 m². S3 writes 24×18 − 22×16 = 80. You have 20 minutes for remediation before the next period, chalk, old newspapers, and a roll of string. A colleague suggests simply teaching the rule 'subtract twice the path width from each dimension'.
1. Diagnose each response. What misconception does each reveal, and what will you do first with each student?
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S1 has the correct structure: outer area minus inner area, and correctly reduces each dimension by 2 m on both sides, giving 20×14. S2 has calculated the perimeter of the field, not an area, then labelled the result in square metres; this is a perimeter–area confusion plus a unit-labelling error. S3 has the right subtractive structure but removes the path width only once from each dimension (22×16), as if the path were on one side only. First moves: with S2, have the student trace the boundary with string and then cover the surface with newspaper squares, saying 'boundary length' and 'surface covered' aloud. With S3, draw the 2 m strip on both sides of one edge and ask him to mark the inner rectangle's corners on the diagram before computing. S1 can be asked to explain the 20 and 14 to a partner, which consolidates the idea without extra worksheets.
2. Design a 20-minute activity using the chalk, string and newspapers that addresses S2 and S3 without simply stating the formula.
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Draw a 24 cm by 18 cm rectangle on the floor or playground with chalk, scaled so 1 cm represents 1 m. Mark a 2 cm strip inside the boundary. Give S2 the string: walk the outer boundary and then the inner boundary, and record both lengths. Ask: 'Does this tell us how much paper covers the path?' Then have the group cover only the strip with torn newspaper pieces, count the approximate area, and compare with the subtraction answer. The physical covering makes 'area' a surface, not a boundary. For S3, fold or mark the strip at a corner and ask him to place a 2 cm ruler on both sides of the same edge; he should see that the inner length is 24 − 2 − 2, not 24 − 2. Finish by having each pair write the two areas, outer and inner, in a table. The activity takes about 12 minutes, leaving 8 for the table and a short whole-class summary.
3. Evaluate the colleague's rule 'subtract twice the path width from each dimension'. When is it valid, and how would you teach it?
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The rule is valid only for a path of uniform width running inside along the whole boundary of a rectangle, because the path reduces each dimension at both ends. It is not a general law: it fails for a path on one side only, for an L-shaped path, or for a path outside the field (where dimensions increase). Teaching it as a remembered formula would let S3 produce 22×16 with confidence, because the rule 'subtract twice' has not been connected to a diagram. Better to teach the diagram first: label the inner rectangle's length as L − 2w and breadth as B − 2w, and only then, if desired, note the shortcut as a summary. For assessment, include one item with a path on one side so that the conditional nature of the rule becomes visible. The rule is a useful compression of reasoning, not a substitute for it.