Numeracy Higher June 2022 Unit 2 Q7
7. Tubu is a company that makes chocolate bars.
The packaging for a new chocolate bar is in the shape of an isosceles triangular prism.
To open the packaging, Tubu is considering adding a tear strip.
The tear strip will go from \(A\) to \(B\), \(B\) to \(C\), and then from \(C\) back to \(A\).
Some of the dimensions of the packaging are shown in the diagram below.
\(DB = DC\).
Diagram not drawn to scale
Calculate the total length of the tear strip required for the packaging. [7]
| Scheme | Marks | Comments |
|---|---|---|
| Strategy of using trigonometry to find DB (or DC) followed by Pythagoras to find AB (or AC) | S1 | Or equivalent full method |
| (DB =) \(\dfrac{3.5}{\cos 65}\) OR (DB =) \(\dfrac{7 \times \sin 65}{\sin 50}\) | M2 | Or a complete method to find DB using the vertical height of the triangle and Pythagoras M1 for cos65 = \(\dfrac{3.5}{DB}\) OR \(\dfrac{DB}{\sin 65} = \dfrac{7}{\sin 50}\) or equivalent |
| = 8.28(1…) or 8.3 (cm) | A1 | CAO Award A0 but FT if e.g. 8 or 8.2 used in next step |
| (AB2 =) 132 + 8.28(1…)2 | M1 | FT ‘their 8.28(1…)’ provided trigonometry attempted to find DB |
| AB2 = 237.58(6…) or 237.6 OR (AB =) \(\sqrt{237.58(6\ldots)}\) or \(\sqrt{237.6}\) or 15.4(138…) (cm) | A1 | FT for similar accuracy Note: use of DB = 8.2 leads to AB2 = 236.24 OR AB = \(\sqrt{236.24}\) or 15.37(01…) use of DB = 8.3 leads to AB2 = 237.89 OR AB = \(\sqrt{237.89}\) or 15.4(236…) |
| (Length of tear strip needed =) 37.8(2…) or 37.83 or 38 (cm) | B1 | FT the correct evaluation of ‘their \(\sqrt{237.58(6\ldots)}\)’ × 2 + 7 provided previous M1 awarded Note: use of DB = 8.2 leads to an answer of 37.7(40…cm) use of DB = 8.3 leads to an answer of 37.8(47…cm) |
Alternative
7. Alternative method:
| Scheme | Marks | Comments |
|---|---|---|
| Strategy of using trigonometry to find the vertical height of the triangle followed by 3-D Pythagoras | S1 | Or equivalent full method |
| (h =) 3.5 × tan65 OR (h =) \(\dfrac{3.5 \times \sin 65}{\sin 25}\) | M2 | M1 for tan65 = \(\dfrac{h}{3.5}\) OR \(\dfrac{h}{\sin 65} = \dfrac{3.5}{\sin 25}\) or equivalent |
| = 7.5(057…) (cm) | A1 | CAO Award A0 but FT if e.g. 7 or 8 used in next step |
| (AB2 =) 7.5(057…)2 + 3.52 + 132 | M1 | FT ‘their 7.5(057…)’ provided trigonometry attempted to find h |
| AB2 = 237.58(6…) or 237.6 OR (AB =) \(\sqrt{237.58(6\ldots)}\) or \(\sqrt{237.6}\) or 15.4(138…) (cm) | A1 | FT for similar accuracy |
| (Length of tear strip needed =) 37.8(2…) or 37.83 or 38 (cm) | B1 | FT the correct evaluation of ‘their \(\sqrt{237.58(6\ldots)}\)’ × 2 + 7 provided previous M1 awarded |