4400 Higher November 2008 Paper 3 Q12
12
Diagram NOT accurately drawn
\(ABC\) and \(AED\) are two straight lines.
\(BE\) is parallel to \(CD\).
\(AE = 5.1\) cm, \(BE = 6\) cm, \(CD = 10\) cm.
(a) Calculate the length of \(DE\). (3)
(b) Calculate the value of \(\dfrac{\text{Area of triangle } ABE}{\text{Area of trapezium } BCDE}\) (3)
| Scheme | Marks |
|---|---|
| \(\frac{10}{6}\) oe or \(\frac{6}{10}\) oe seen | B1 |
| \(5.1 \times \frac{10}{6}\) or \(5.1 \div \frac{6}{10}\) or 8.5 | M1 |
| 3.4 | A1 |
| (3) |
Notes
B1: for \(\frac{10}{6}\) oe (1.666...) or \(\frac{6}{10}\) oe (0.6) or \(\frac{2}{3}\) (0.666...)
M1: for \(5.1 \times \frac{10}{6}\) or \(5.1 \div \frac{6}{10}\) or \(5.1 \times \frac{2}{3}\) or 8.5
A1: cao
| Scheme | Marks |
|---|---|
(scale factor)\(^2\) eg \(\left(\frac{6}{10}\right)^2\) or \(\frac{36}{100}\) or \(\left(\frac{10}{6}\right)^2\) or \(\frac{100}{36}\) | M1 |
| eg \(100 - 36\), 64, \(1 - \frac{36}{100}\), \(\frac{64}{100}\) | M1 |
| \(\frac{9}{16}\) oe | A1 |
| (3) | |
| (6 marks) |
Notes
M1: M2 for \(\dfrac{\frac{1}{2} \times 6 \times 5.1 \sin\theta}{\frac{1}{2} \times (10 + 6) \times 3.4 \sin\theta}\) or \(\dfrac{\frac{1}{2} \times 6 \times 5.1 \sin\theta}{\frac{1}{2} \times 10 \times 8.5 \sin\theta - \frac{1}{2} \times 6 \times 5.1 \sin\theta}\)