Numeracy Intermediate June 2018 Unit 2 Q9
9. Yared is going to make a door wedge.
The cross-section of the wedge is shown below.
The horizontal length is 12 cm and the vertical height is 3 cm.

Diagram not drawn to scale
Calculate the length \(x\).
Give your answer correct to 3 significant figures. [4]
\(x\) = …………… cm
The wedge must fit under Yared’s door.
The angle \(y\) must be less than 15°.
Show that this wedge will fit under Yared’s door.
You must show all your working. [3]
Yared decides to make a larger wedge that is mathematically similar to the one shown in part (a).
This wedge is to have a vertical height of 4·5 cm.

Diagram not drawn to scale
Calculate the horizontal length of this door wedge. [2]
The wedge will be …………… cm long
| Scheme | Marks | Comments |
|---|---|---|
| (i) (x2 =) 32 + 122 | M1 | Or alternative full method |
| x2 = 153 or (x =) √153 | A1 | Or accurate first stage of alternative full method |
| 12.4 (cm) | A2 | Must be 3 s.f. A1 for sight of (x =) 12.40 or 12.3(69…cm), NOT for 12 (cm) FT from M1 for the correctly evaluated square root of ‘their 153’ provided ‘their answer’ > 12 (cm) |
| (ii) (y=) tan-13/12 or tan-1 0.25, or cos-1 12/12.4 or cos-1 0.967.. or cos-1 0.97.., or sin-13/12.4 or sin-10.24…, | M2 | FT ‘their 12.4’ provided > 12 (cm) M1 for tany = 3/12 or cosy = 12/12.4 or siny = 3/12.4 |
| (y =)14.(0...°) (which is < 15°) | A1 | Accept 14.(59...°) from use of 12.4 cm |
| Scheme | Marks | Comments |
|---|---|---|
| 12 × 4.5 ÷ 3 or 4.5/tan14(.0..°) | M1 | FT ‘their angle y’ |
| 18(cm) or 18.0(… cm) | A1 | ISW to calculate the hypotenuse |