Higher November 2020 Paper 1R Q17
17 The diagram shows a prism \(ABCDEFGH\) with a horizontal base.
Diagram NOT accurately drawn
The base of the prism, \(EFGH\), is a square of side 12 cm.
Trapezium \(ADEF\) is a cross section of the prism where \(AF\) and \(DE\) are vertical edges.
\(DE = CH = 10\) cm
\(AD = BC = 15\) cm
(a) Work out the size of the angle between \(CF\) and the base \(EFGH\).
Give your answer correct to one decimal place. (3)
Give your answer correct to one decimal place. (3)
(b) Work out the length of \(BE\).
Give your answer correct to one decimal place. (3)
Give your answer correct to one decimal place. (3)
| Scheme | Marks |
|---|---|
| \((FH =)\; \sqrt{12^2 + 12^2}\;(= 16.97\ldots \text{ or } \sqrt{288} \text{ or } 12\sqrt{2})\) | M1 |
\(\tan CFH = \dfrac{10}{\text{“}{16.97\ldots}\text{”}}\) oe or e.g. \((CF =)\; \sqrt{\text{“}{16.97}\text{”}^2 + 10^2}\;(= 19.69\ldots \text{ or } \sqrt{388} \text{ or } 2\sqrt{97})\) and e.g. \(\dfrac{\sin CFH}{10} = \dfrac{\sin 90}{\text{“}{19.69}\text{”}}\) | M1 |
| 30.5 | A1 |
| (3) |
Notes
M1: for a correct trig statement involving \(CFH\)
A1: accept 30.4 – 30.7
| Scheme | Marks |
|---|---|
| \((BG =)\; 10 + \sqrt{15^2 - 12^2}\;\;(= 19)\) | M1 |
| \((BE =)\; \sqrt{\text{“}{19}\text{”}^2 + \text{“}{16.97\ldots}\text{”}^2}\) oe | M1 |
| 25.5 | A1 |
| (3) | |
| (6 marks) |
Notes
M1: ft their \(FH\)
A1: accept 25.4 – 25.6