Numeracy Higher November 2016 Unit 2 Q8
8. A confectionary company is designing a new chocolate-covered biscuit in the shape of a square-based pyramid.
The centre of the square base is labelled \(O\).
Each biscuit will have base sides of length 3·4 cm, and a vertical height of 2·1 cm.
Diagram not drawn to scale
(a) Calculate the angle that one of the triangular faces makes with the base of the pyramid. [4]
(b) The company knows that it costs 0·08p per cm2 to apply a chocolate covering.
Calculate the cost of applying a chocolate covering to all 5 faces of a biscuit. [6]
Calculate the cost of applying a chocolate covering to all 5 faces of a biscuit. [6]
| Scheme | Marks | Comments |
|---|---|---|
| Identifying the length 1·7 on a correct triangle | S1 | May be seen or implied If (a) is unanswered, may be awarded from (b) |
| (Angle =) tan−1(2·1/1·7) | M2 | If hypotenuse used, a complete method needed M1 for tan(angle) = 2·1/1·7 |
| = 51(·009…)(°) | A1 | Allow an answer that rounds to 51(°) |
| Scheme | Marks | Comments |
|---|---|---|
| (Slant height of triangular face2 =) 2·12 + 1·72 Height2 = 7·3 (Height =) 2·7(018…) OR √7·3 (cm) | M1 A1 A1 | FT from M1 for the correctly evaluated square root of ‘their 7·3’ provided ‘their answer’ > 2·1 (cm) Alternative methods: M2 for (Height =) 2·1/sin 51(·009…)(°) OR M2 for (Height =) 1·7/cos 51(·009…)(°) FT ‘their derived 51(·009…)(°)’ M1 for sin 51(·009…)(°) = 2·1/Height OR M1 for cos 51(·009…)(°) = 1·7/Height A1 for 2·7(018…) (cm) |
| (Total s. area =) 4×½×3·4×2·7(018…) + 3·42 OR 4×½×3·4×√7·3 + 3·42 = 29·9(325…)(cm2) | M1 A1 | FT ‘their derived 2·7(018…)’ Note: 3·42 = 11·56 |
| (Cost of chocolate =) 2·39(46…) or 2·4 (p) | B1 | FT 0·08בtheir 29·9(325…)’ provided at least one of the previous M1 marks awarded Allow an answer of 2 (p) from correct working If units are given, they must be correct |