(a) 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
(i) Calculate the length \(x\).
Give your answer correct to 3 significant figures. [4]
\(x\) = …………………… cm
(ii) 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]
(b) 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
Mark scheme (a)
Scheme
Marks
Comments
(i) (x2 =) 32 + 122 x2 = 153 or (x =) √153
M1 A1
Or alternative full method 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
Mark scheme (b)
Scheme
Marks
Comments
12 × 4.5 ÷ 3 or 4.5/tan14(.0..°) 18(cm) or 18.0(… cm)
M1 A1
FT ‘their angle y’ ISW to calculate the hypotenuse
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