Higher June 2017 Unit 2 Q14
14. The region between two rectangles is shaded, as shown in the diagram below.
All of the measurements shown are given correct to the nearest cm.
Diagram not drawn to scale
Calculate the greatest possible area of the shaded region. [3]
| Scheme | Marks | Comments |
|---|---|---|
| (Greatest area =) 31.5 x 23.5 – 20.5 x 12.5 (= 740.25 – 256.25) | M2 | Award M1 for correct use of values 31< l ≤ 31.5, 23<w≤23.5, 20.5≤ l < 21, 12.5≤w<13. OR M1 for 31.5 x 23.5 – ‘area of inner rectangle’ OR M1 for ‘area of outer rectangle’ – 20.5 x 12.5 |
| = 484 (cm2) | A1 | CAO |
Alternative examples for method marks
(adding up split areas of the shaded region).
| Scheme | Marks | Comments |
|---|---|---|
| 1. Horizontal split 2 x 31.5 x 5.5 + 2 x 12.5 x 5.5 , | M2 | |
| OR 2. Vertical split 2 x 23.5 x 5.5 + 2 x 20.5 x 5.5 , | M2 |
Award M1 for correct use of values 31< l ≤ 31.5, 23<w≤23.5, 20.5≤ l < 21, 12.5≤w<13 and ‘their 5.5’ adjusted accordingly to their values.
Note that the ‘shaded width’ need not be consistent around the inner rectangle.