Higher January 2021 Paper 2 Q17
17 The diagram shows a solid prism \(ABCDEFGH\).
Diagram NOT accurately drawn
The trapezium \(ABCD\), in which \(AD\) is parallel to \(BC\), is a cross section of the prism.
The base \(ADEH\) of the prism is a horizontal plane.
\(ADEH\) and \(BCFG\) are rectangles.
The midpoint of \(BC\) is vertically above the midpoint of \(AD\) so that \(BA = CD\).
\(AD\) = 37 cm \(GF\) = 28 cm \(DE\) = 24 cm
The perpendicular distance between edges \(AD\) and \(BC\) is 20 cm.
Give your answer correct to one decimal place. (3)
| Scheme | Marks |
|---|---|
| \(\left(\dfrac{37 + 28}{2}\right) \times 20\) (= 650) | M1 |
| \(\sqrt{4.5^2 + 20^2}\) (= 20.5) oe | M1 |
| 2 × ‘650’ + 2 × ‘20.5’ × 24 + 37 × 24 + 28 × 24 (2 × ‘650’ + 2 × 492 + 888 + 672) | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 3844 | A1 |
| (4) |
Notes
M1: Correct method to find area of trapezium
M1: Correct method to find slanted edge \(AB\) oe
M1: method to find the sum of the surface areas of at least 4 correct faces (ft their area of trapezium) ignore incorrect areas
| Scheme | Marks |
|---|---|
eg \(\sqrt{24^2 + (37 - \text{“}4.5\text{”})^2}\) (= 40.4) \((AF =)\;\sqrt{24^2 + 20^2 + (37 - \text{“}4.5\text{”})^2}\) (= 45.08...) | M1 |
\(\tan x = \dfrac{20}{\text{“}40.4\text{”}}\) or \(\sin x = \dfrac{20(\sin 90)}{\text{“}45.08\text{”}}\) or \(\cos x = \dfrac{\text{“}40.4\text{”}^2 + \text{“}45.08\text{”}^2 - 20^2}{2 \times \text{“}40.4\text{”} \times \text{“}45.08\text{”}}\) | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 26.3 | A1 |
| (3) | |
| (7 marks) |
Notes
M1: Correct method to find diagonal from \(A\) to point on \(HE\) below \(F\) or \(AF\)
M1: Correct trig statement for finding the required angle
A1: 26.3 – 26.4