Higher June 2024 Paper 6 Q9
9 The diagram shows a shaded sector inside a parallelogram.
The sector has an angle of 30°.
The parallelogram, ABCD, has length BC = 20 cm and AB = 12 cm.
The perpendicular distance between BC and AD is 6 cm.
Not to scale
(a) Show that the area of the sector is \(37.7\,\text{cm}^2\), correct to 3 significant figures. [3]
(b) Work out the percentage of the parallelogram that is not shaded. [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\frac{30}{360} \times \pi \times 12^2\) oe | M2 | M1 for \(\pi \times 12^2\) implied by \(144\pi\) or 452.3 to 452.45 | M2 oe e.g. \(\frac{360}{30} = 12\) and \(\frac{\pi \times 12^2}{12}\) Condone 3.14 or \(\frac{22}{7}\) for \(\pi\) in M marks M0 for \(12\pi\) without working |
| 37.69 to 37.704 | A1 | A0 for just 37.7 | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 68.5 to 68.6 nfww | 4 | With area of parallelogram as 120 or from an attempt at 20 × 6: | If 120 not used, their (20 × 6) must come from attempt at 20 × 6 Note there are other methods for finding area of a parallelogram 37.7 may be their more accurate 37.69 to 37.704 from (a) |
| M3 for \(\frac{\textit{their}(20 \times 6) - 37.7}{\textit{their}(20 \times 6)}\) [× 100] oe or M2 for their (20 × 6) – 37.7 implied by 82.3 or for \(\frac{37.7}{\textit{their}(20 \times 6)}\) [× 100] implied by 31.4 to 31.5 or M1 for 20 × 6 implied by 120 | M3 oe e.g. \(100 - \frac{37.7 \times 100}{\textit{their}(20 \times 6)}\) | ||
| If correct method for area of parallelogram is not shown or an incorrect value is used: If \(A \gt 37.7\) SC2 for \(\frac{A - 37.7}{A}\) [× 100] oe or SC1 for \(\frac{37.7}{A}\) [× 100] | e.g SC2 for \(\frac{240 - 37.7}{240}\) [× 100] oe For SC marks method must be shown; do not imply from an answer only | ||