P6 Maths Composite Shapes Paper 1: Basics
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: P6 Maths Composite Shapes with Circles |
| Level | P6 |
| # | Answer | One-line note |
|---|---|---|
| 1 | 78.5 cm² | The largest circle has diameter equal to the side, so the radius is 5 cm and 3.14 × 5 × 5 = 78.5 cm². |
| 2 | 8 cm | The diameter is limited by the shorter side, so it is 8 cm. |
| 3 | 28.26 cm² | 3.14 × 6 × 6 ÷ 4 = 28.26 cm². |
| 4 | 21.5 cm² | The square is 100 cm² and the circle is 78.5 cm², so 21.5 cm² remains. |
| 5 | 25.7 cm | The arc is 3.14 × 10 ÷ 2 = 15.7 cm, and adding the 10 cm diameter gives 25.7 cm. |
| 6 | 121.5 cm² | The two semicircles make one circle of 78.5 cm², so 200 − 78.5 = 121.5 cm² remains. |
| 7 | 49.98 cm | The arc is 2 × 3.14 × 14 ÷ 4 = 21.98 cm, and two radii add 28 cm, giving 49.98 cm. |
| 8 | 200.96 cm² | 3.14 × (100 − 36) = 200.96 cm². |
| 9 | 2355 dollars | The area is 3.14 × 100 ÷ 2 = 157 m², so 157 × 15 = 2355 dollars. |
| 10 | 30.84 cm | The perimeter of a semicircle includes the diameter: 18.84 + 12 = 30.84 cm. |
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