Higher June 2019 Paper 2 Q7
7 The diagram shows rectangle \(ABDE\) and right-angled triangle \(ABC\).
\(AC = 17\) cm
\(BC = 8\) cm
Not drawn accurately
\(BC : CD = 1 : 2\)
Work out the area of rectangle \(ABDE\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(8^2\) or 64 and \(17^2\) or 289 | M1 | |
| \(\sqrt{17^2 - 8^2}\) or \(\sqrt{225}\) or 15 | M1dep | oe implies M2 may be seen on diagram |
| \(8 \times 3 \times\) their 15 or \(24 \times\) their 15 | M1dep | dep on M2 oe eg \((8 + 16) \times\) their 15 or \(0.5 \times 8 \times\) their \(15 \times 6\) |
| 360 | A1 | SC2 [448.8, 456] |
| Alternative method 2 | ||
| \(\cos C = \dfrac{8}{17}\) or \(C = [61.9, 62]\) | M1 | may be seen on diagram |
| \(17 \times \sin\) their [61.9, 62] or [14.9, 15.1] | M1dep | may be seen on diagram oe eg \(8 \times \tan\) their [61.9, 62] |
| \(8 \times 3 \times\) their [14.9, 15.1] or \(24 \times\) their [14.9, 15.1] or [357.6, 362.4] | M1dep | dep on M2 oe eg \((8 + 16) \times\) their [14.9, 15.1] or \(0.5 \times 8 \times\) their [14.9, 15.1] \(\times\ 6\) |
| 360 | A1 | SC2 [448.8, 456] |
| Alternative method 3 | ||
| \(\sin A = \dfrac{8}{17}\) or \(A = [28, 28.1]\) | M1 | may be seen on diagram |
| \(17 \times \cos\) their [28, 28.1] or [14.9, 15.1] | M1dep | may be seen on diagram oe eg \(8 \div \tan\) their [28, 28.1] |
| \(8 \times 3 \times\) their [14.9, 15.1] or \(24 \times\) their [14.9, 15.1] or [357.6, 362.4] | M1dep | dep on M2 oe eg \((8 + 16) \times\) their [14.9, 15.1] or \(0.5 \times 8 \times\) their [14.9, 15.1] \(\times\ 6\) |
| 360 | A1 | SC2 [448.8, 456] |
| Alternative method 4 | ||
| \(\cos C = \dfrac{8}{17}\) or \(C = [61.9, 62]\) | M1 | may be seen on diagram |
| \(\dfrac{1}{2} \times 8 \times 17 \times \sin\) their [61.9, 62] or [59.9, 60.1] | M1dep | oe |
| \(6 \times\) their [59.9, 60.1] or [357.6, 362.4] | M1dep | oe |
| 360 | A1 | SC2 [448.8, 456] |
Additional guidance
| 15 without a contradictory value for \(AB\) scores the first two marks on Alt method 1, even if not subsequently used | M1M1 |
| \(\sqrt{17^2 + 8^2}\) | M1M0 |
| 3rd M1 is for the total area and may be calculated in various ways eg using a trapezium + a triangle | |
| 3rd M1 is for the total area so further work will lose the mark eg 360 seen followed by 360 – 60, answer 300 | M1M1M0A0 |
| May use sine rule or cosine rule but must reach \(AB = \ldots\) to award the second M1 in Alt 2 or 3 |