P4 Maths Perimeter and Area Applications Paper 4: Error Review
Ten original items, each with an answer and a one-line explanation.
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 20 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: P4 Maths Perimeter and Area Applications |
| Level | P4 |
| # | Answer | One-line note |
|---|---|---|
| 1 | 38 cm | Perimeter = (length + width) × 2 = (12 + 7) × 2 = 38 cm. |
| 2 | 81 cm² | Area of a square = side × side = 9 × 9 = 81 cm². |
| 3 | 8 cm | Length = area ÷ width = 48 ÷ 6 = 8 cm, and 8 × 6 = 48 checks it. |
| 4 | 81 cm² | The side is 36 ÷ 4 = 9 cm, so the area is 9 × 9 = 81 cm². |
| 5 | 60000 cm² | The area is 6 m², and 1 m² = 10 000 cm², so 6 × 10 000 = 60000 cm². |
| 6 | 76 cm² | 12 × 8 = 96 and 5 × 4 = 20, so the area is 96 − 20 = 76 cm². |
| 7 | 36 cm² | The wire is the perimeter, so each side is 24 ÷ 4 = 6 cm and the area is 36 cm². |
| 8 | 23 trees | The perimeter is 46 m and 46 ÷ 2 = 23; going round a closed loop, trees equal gaps. |
| 9 | 12 cm² | The area goes from 6 × 4 = 24 cm² to 9 × 4 = 36 cm², an increase of 12 cm². |
| 10 | No: shapes with the same area can have very different perimeters | A 4 cm by 4 cm square has area 16 cm² and perimeter 16 cm; a 1 cm by 16 cm rectangle has the same area but a perimeter of 34 cm. |
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