Higher November 2024 Paper 2 Q6
6 Here is a right-angled triangle.
Not drawn accurately
Use Pythagoras’ theorem to show that \(\quad x = 0.8\) [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(1.7^2 - 1.5^2 = 0.64\) and \(\sqrt{0.64} = 0.8\) or \(2.89 - 2.25 = 0.64\) and \(\sqrt{0.64} = 0.8\) | B2 | accept \(0.8^2 = 0.64\) for \(\sqrt{0.64} = 0.8\) accept \(\sqrt{1.7^2 - 1.5^2} = 0.8\) for B2 accept \(1.7^2 - 1.5^2 = 0.8^2\) for B2 B1 \(1.7^2\) and \(1.5^2\) oe |
| Alternative method 2 | ||
| \(1.7^2 - 0.8^2 = 2.25\) and \(\sqrt{2.25} = 1.5\) or \(2.89 - 0.64 = 2.25\) and \(\sqrt{2.25} = 1.5\) | B2 | accept \(1.5^2 = 2.25\) for \(\sqrt{2.25} = 1.5\) accept \(\sqrt{1.7^2 - 0.8^2} = 1.5\) for B2 accept \(1.7^2 - 0.8^2 = 1.5^2\) for B2 B1 \(1.7^2\) and \(0.8^2\) oe |
| Alternative method 3 | ||
| \(0.8^2 + 1.5^2 = 2.89\) and \(\sqrt{2.89} = 1.7\) or \(0.64 + 2.25 = 2.89\) and \(\sqrt{2.89} = 1.7\) | B2 | accept \(1.7^2 = 2.89\) for \(\sqrt{2.89} = 1.7\) accept \(\sqrt{0.8^2 + 1.5^2} = 1.7\) for B2 accept \(0.8^2 + 1.5^2 = 1.7^2\) for B2 B1 \(0.8^2\) and \(1.5^2\) oe |
Additional guidance
| B1 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts | |
| \(1.7^2 - 1.5^2 = 0.64 \quad x^2 = 0.64 \quad x = 0.8\) | B2 |
| Max B1 if any incorrect statement seen eg \(1.7^2 - 1.5^2 = \sqrt{0.64} = 0.8\) | B1 |
| Accept \(1.7 \times 1.7\) for \(1.7^2\) etc | |
| Condone eg 1.5 cm\(^2\) and 1.7 cm\(^2\) for \(1.5^2\) and \(1.7^2\) for B1 but must be recovered for B2 | |
| \(0.64 \div 0.8 = 0.8\) is equivalent to \(\sqrt{0.64} = 0.8\) |