S2 Maths Pythagoras Theorem Paper 2: Intermediate
Diagonals, ladders, coordinate distances and isosceles triangles.
| 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 | 20 cm | 144 + 256 = 400, so the diagonal is 20 cm. |
| 2 | 40 cm | 1681 − 81 = 1600, so the other leg is 40 cm. |
| 3 | 3 m | 5.76 + 3.24 = 9, so the ladder is 3 m long. |
| 4 | 29 cm | 400 + 441 = 841, so the hypotenuse is 29 cm. |
| 5 | 24 cm | 676 − 100 = 576, so the other leg is 24 cm. |
| 6 | 15 units | The differences are 9 and 12; 81 + 144 = 225, so AB = 15 units. |
| 7 | Yes | 64 + 225 = 289, which equals 17², so it is right-angled. |
| 8 | 10 cm | Half the base is 8; 64 + 36 = 100, so each equal side is 10 cm. |
| 9 | No | The hypotenuse is the longest side, so 8 m cannot be the hypotenuse when a leg is 15 m. |
| 10 | Subtract instead: 169 − 25 = 144, so the other leg is 12 cm | To find a leg, subtract the known leg squared from the hypotenuse squared. |
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