Foundation June 2025 Paper 3 Q23
23
Not drawn accurately
Use Pythagoras’ theorem to show that the value of \(x\) is between 10 and 11 [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(12^2\) or \(13^2\) | M1 | oe 144 or 169 implied by 313 or \(\sqrt{313}\) or [17.6, 17.7] |
| \(13^2 - 12^2\) or \(169 - 144\) or 25 or \(\sqrt{13^2 - 12^2}\) or \(\sqrt{169 - 144}\) or \(\sqrt{25}\) or 5 | M1dep | oe 5 may be in correct position on the diagram |
| \(9^2 + (\text{their } 5)^2\) or \(81 +\) their 25 or 106 or \(\sqrt{9^2 + (\text{their } 5)^2}\) or \(\sqrt{81 + \text{their } 25}\) or \(\sqrt{106}\) or [10.2, 10.3] | M1dep | oe their 5 or their 25 must be from correct working |
| \(9^2 + 5^2 = 106\) and \(\sqrt{106}\) and [10.2, 10.3] or \(\sqrt{9^2 + 5^2}\) and [10.2, 10.3] or \(\sqrt{81 + 25}\) and [10.2, 10.3] or \(9^2 + 5^2 = 106\) and \(10^2 = 100\) and \(11^2 = 121\) | A1 | oe |
Additional guidance
| Up to M2 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts | |
| Condone eg 12 cm\(^2\) for \(12^2\) | |
| \(9^2 + 5^2 = 106 \quad x^2 = 106 \quad x = 10.3\) | M3A1 |
| Ignore further working after correct answer eg \(\sqrt{9^2 + 5^2}\) and 10.3 with \(10 \gt 10.3 \lt 11\) | M3A1 |
| \(9^2 + 5^2 = \sqrt{106} = 10.3\) omission of \(\sqrt{9^2 + 5^2}\) or \(9^2 + 5^2 = 106\) is a missing step | M3A0 |
| Using trigonometry or accurate/scale drawing only | M0 |