S2 Maths Pythagoras Theorem Paper 3: Challenge
Space diagonals, the converse and reverse problems needing several steps.
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 25 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S2 Maths: Pythagoras Theorem |
| Level | S2 |
| # | Answer | One-line note |
|---|---|---|
| 1 | 13 cm | 9 + 16 + 144 = 169, so the space diagonal is 13 cm. |
| 2 | 10 units | The differences are 6 and 8; 36 + 64 = 100, so PQ = 10 units. |
| 3 | 90 cm | The hypotenuse is √1681 = 41, so the perimeter is 9 + 40 + 41 = 90 cm. |
| 4 | 12 m | 225 − 81 = 144, so the top reaches 12 m. |
| 5 | Yes | 2.25 + 4 = 6.25, which equals 2.5², so it is right-angled. |
| 6 | 10 cm | Twice the square of the side is 200, so the side squared is 100 and the side is 10 cm. |
| 7 | 120 cm² | 100 + 576 = 676, so the right angle is between 10 and 24; area = 10 × 24 ÷ 2 = 120 cm². |
| 8 | 150 cm² | The legs are 15 cm and 20 cm, so the area is 15 × 20 ÷ 2 = 150 cm². |
| 9 | No | 49 + 81 = 130, which is less than 144, so it is not right-angled. |
| 10 | 5.66 cm | Each leg is 8 ÷ √2 = 5.66 cm; halving the hypotenuse is wrong. |
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