Numeracy Higher June 2019 Unit 1 Q10
10. A tent company is designing a new 2-person tent.
The base of the tent is in the shape of a kite, as shown below.
The width of the kite is 160 cm, and the two shorter sides are of length 100 cm.
The point where the diagonals of the kite intersect has been marked \(O\) on the diagram below.
Diagram not drawn to scale
\(E\) is the highest point of the tent, and is 110 cm vertically above \(O\).
Part of the frame that supports the tent cover is a straight pole that goes from \(A\) to \(E\).
Diagram not drawn to scale
Calculate the length of pole \(AE\).
Give your answer as a surd.
You do not need to simplify your answer. [4]
| Scheme | Marks | Comments |
|---|---|---|
| AO2 = 1002 − 802 or (AO =) \(\sqrt{100^2 - 80^2}\) | M1 | Allow use of 20 × √(52 − 42) |
| AO2 = 3600 or (AO =) √3600 or (AO =) 60 (cm) | A1 | If √3600 evaluated, mark final answer |
| AE2 = 1102 + 3600 or equivalent OR (AE =) \(\sqrt{110^2 + 60^2}\) or equivalent | M1 | FT ‘their derived 3600’ or ‘their derived 60’ but not use of 100 or 80 for AO |
| AE = √15700 (cm) ISW | A1 | Or 10√157 |
Alternative method:
| Scheme | Marks | Comments |
|---|---|---|
| AO2 = 1002 − 802 or (AO =) \(\sqrt{100^2 - 80^2}\) | M1 | |
| (AE =) \(\sqrt{100^2 - 80^2 + 110^2}\) | M2 | M1 for 1002 − 802 + 1102 |
| AE = √15700 (cm) ISW | A1 | Or 10√157 If no marks awarded, SC2 for \(\sqrt{100^2 + 80^2 + 110^2}\) leading to an answer of \(\sqrt{28500}\) ISW, or SC1 for 1002 + 802 + 1102 |