Numeracy Higher November 2022 Unit 2 Q7
7.
The Great Pyramid of Giza, in Egypt, is the oldest of the Seven Wonders of the Ancient World.
It is a square-based pyramid of base length 230 m.
It stands on horizontal ground.
Each of the sloping edges of the pyramid has a length of 217 m.
Diagram not drawn to scale
Calculate the volume of the Great Pyramid of Giza. [7]
A gift shop in Giza makes souvenir pyramids.
They are based on the Great Pyramid.
The shop paints all the faces of the souvenir pyramids.
The base length of a souvenir pyramid is related to the total surface area to be painted by the formula:
where:
- \(b\) is the base length
- \(A\) is the total surface area to be painted.
Calculate the total surface area to be painted of a souvenir pyramid that has a base length of 12 cm. [3]
The Sphinx is another ancient monument that sits next to the Great Pyramid.
The shop sells two sizes of Sphinx souvenirs that are mathematically similar.
Diagrams not drawn to scale
The small souvenir has a length of 15 cm and a total surface area to be painted of 400 cm\(^2\). The large souvenir has a length of 31·5 cm.
Calculate the total surface area to be painted of the large souvenir. [3]
| Scheme | Marks | Comments |
|---|---|---|
| Strategy of using Pythagoras in 2 different planes to calculate the vertical height | S1 | |
| 115\(^2\) + 115\(^2\) OR \(\dfrac{230^2 + 230^2}{4}\) OR 217\(^2\) − 115\(^2\) | M1 | Or their square roots Note: 115\(^2\) + 115\(^2\) and \(\dfrac{230^2 + 230^2}{4}\) = 26 450, and \(\sqrt{115^2 + 115^2}\) and \(\dfrac{\sqrt{230^2 + 230^2}}{2}\) = 162.6(3…) |
| (Vertical height =) \(\sqrt{217^2 - (115^2 + 115^2)}\) OR \(\sqrt{217^2 - \dfrac{230^2 + 230^2}{4}}\) (=\(\sqrt{20639}\)) | M2 | Awarding of M2 or M1 here implies previous S1M1 M1 for 217\(^2\) – (115\(^2\) + 115\(^2\)) or M1 for 217\(^2\) – \(\dfrac{230^2 + 230^2}{4}\) or equivalent, or M1 for 217\(^2\) = h\(^2\) + (115\(^2\) + 115\(^2\)) or M1 for 217\(^2\) = h\(^2\) + \(\dfrac{230^2 + 230^2}{4}\) or equivalent |
| = 143.6(627…) to 143.7 (m) | A1 | Allow 144 (m) provided no incorrect work seen |
| (Volume of pyramid =) \(\dfrac{1}{3}\) × 230 × 230 × 143.6(627…) | M1 | FT ‘their derived 143.6(627…)’ |
| = 2 533 254(.034) (m\(^3\)) | A1 | Allow answers of 2530000 to 2534000 A height of:
|
(i)
| Scheme | Marks | Comments |
|---|---|---|
| \(\dfrac{A}{1+\tan 58} = b^2\) OR \(\dfrac{A}{1+\tan 58} = 12^2\) OR \(\dfrac{A}{1+\tan 58} = 144\) | B1 | Note: 1 + tan58(º) = 2.6(00334…) |
| \((A =)\ b^2(1+\tan 58)\) OR \((A =)\ 12^2(1+\tan 58)\) OR \((A =)\ 144(1+\tan 58)\) | B1 | Implies previous B1 |
| (A =) 374.4(481…) (cm\(^2\)) | B1 | Implies previous B1B1 |
(ii)
| Scheme | Marks | Comments |
|---|---|---|
| (Area factor =) \(\left(\dfrac{31.5}{15}\right)^2\) OR \(\left(\dfrac{15}{31.5}\right)^2\) or 2.1\(^2\) OR 0.476…\(^2\) (= 4.41) (= 0.2267…) | B1 | May be implied in further working |
| (Area of large souvenir to be painted =) 400 × \(\left(\dfrac{31.5}{15}\right)^2\) OR 400 ÷ \(\left(\dfrac{15}{31.5}\right)^2\) | M1 | |
| = 1764 (cm\(^2\)) | A1 |