June 2019 Paper 2 Q19
19 \(ABC\) is a right-angled triangle.
\(ACD\) is an isosceles triangle.
All dimensions are in centimetres.
(a) Show that \(\quad AC = 5x\) [1 mark]
(b) Work out an expression, in cm2, for the area of quadrilateral \(ABCD\).
Give your answer in the form \(\quad px^2 \quad\) where \(p\) is an integer. [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\sqrt{(3x)^2 + (4x)^2}\ (= 5x)\) or \(\sqrt{9x^2 + 16x^2}\ (= 5x)\) or \((3x)^2 + (4x)^2 = (5x)^2\) or \(3x\), \(4x\), \(5x\) triangle | B1 | may be seen in stages eg \(9x^2 + 16x^2 = 25x^2\) and \(\sqrt{25x^2}\ (= 5x)\) |
Additional guidance
| Only \(\sqrt{25x^2}\ (= 5x)\) seen | B0 |
| Pythagorean triple \(3x\), \(4x\), \(5x\) | B1 |
| Pythagorean triple 3, 4, 5 | B0 |
| Missing brackets can not be recovered eg1 \(\ \sqrt{3x^2 + 4x^2} = 5x\) eg2 \(\ 3x^2 + 4x^2 = 9x^2 + 16x^2 = 25x^2\) and \(\sqrt{25x^2}\ (= 5x)\) | B0 B0 |
| Incorrect statements are B0 (mark the full response) eg1 \(\ 9x^2 + 16x^2 = 25x^2 = \sqrt{25x^2}\ (= 5x)\) eg2 \(\ 9x^2 + 16x^2 = 25x^2 \quad \sqrt{25}\,x^2\ (= 5x)\) eg3 \(\ \sqrt{(3x)^2 + (4x)^2} = 5x\) and \(3x + 4x = 5x\) | B0 B0 B0 |
| Only uses values for \(x\) | B0 |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(0.5 \times 4x \times 3x\) or \(6x^2\) | M1 | oe may be seen on the diagram |
| \((6.5x)^2 - (2.5x)^2\) or \(42.25x^2 - 6.25x^2\) or \(36x^2\) | M1 | oe eg \((6.5x)^2 - \left(\dfrac{5x}{2}\right)^2\) |
| \(\sqrt{\text{their } 36x^2}\) or \(6x\) | M1dep | dep on 2nd M1 may be seen on the diagram |
| \(0.5 \times 5x \times\) their \(6x\) or \(15x^2\) | M1dep | oe dep on 2nd and 3rd M1 |
| \(21x^2\) | A1 | allow \(p = 21\) if areas \(6x^2\) and \(15x^2\) seen |
| Alternative method 2 | ||
| \(0.5 \times 4x \times 3x\) or \(6x^2\) | M1 | oe may be seen on the diagram |
| \(\cos ACD = \dfrac{2.5x}{6.5x}\) or \(\cos ACD = \dfrac{5}{13}\) | M1 | oe |
| \(\cos^{-1} \dfrac{2.5x}{6.5x}\) or 67(.3…) or 67.4 | M1dep | oe eg \(\cos^{-1} \dfrac{(6.5x)^2 + (5x)^2 - (6.5x)^2}{2 \times 6.5x \times 5x}\) dep on 2nd M1 |
| \(0.5 \times 5x \times 6.5x \times \sin\) their 67(.3…) or \(15x^2\) | M1dep | oe dep on 2nd and 3rd M1 |
| \(21x^2\) | A1 | allow \(p = 21\) if areas \(6x^2\) and \(15x^2\) seen |
| Alternative method 3 | ||
| \(0.5 \times 4x \times 3x\) or \(6x^2\) | M1 | oe may be seen on the diagram |
| \((5x)^2 = (6.5x)^2 + (6.5x)^2\) \(- 2 \times 6.5x \times 6.5x \times \cos D\) | M1 | oe |
| \(\cos^{-1} \dfrac{(6.5x)^2 + (6.5x)^2 - (5x)^2}{2 \times 6.5x \times 6.5x}\) or \(\cos^{-1} \dfrac{119}{169}\) or 45(.2…) | M1dep | oe dep on 2nd M1 |
| \(0.5 \times 6.5x \times 6.5x \times \sin\) their 45(.2…) or \(15x^2\) | M1dep | oe dep on 2nd and 3rd M1 |
| \(21x^2\) | A1 | allow \(p = 21\) if areas \(6x^2\) and \(15x^2\) seen |
Additional guidance
| Allow recovery of algebra eg1 \(\ 0.5 \times 4 \times 3 = 6\) is 1st M0 but if recovered to \(6x^2\) scores 1st M1 eg2 Alt 1 \(\sqrt{42.25 - 6.25} = 6\) is 2nd M0 and 3rd M0 but if recovered to \(6x\) scores 2nd M1 and 3rd M1 | |
| Do not allow final mark if an incorrect area is seen eg do not allow answer \(21x^2\) if their two areas are \(6x^2\) and \(15x\) | |
| Answer \(21x^2\) with no incorrect working eg fully correct working with numbers and final answer \(21x^2\) | 5 marks |
| Allow recovery of missing brackets | |
| Choose the scheme that favours the student |