Higher June 2019 Paper 2R Q23
23 The diagram shows a solid pyramid \(ABCDE\) with a horizontal base.
Diagram NOT accurately drawn
The base, \(BCDE\), of the pyramid is a square of side 10 cm.
The vertex \(A\) of the pyramid is vertically above the centre \(O\) of the base so that \(AB = AC = AD = AE\)
The total surface area of the pyramid is 360 cm2
Work out the size of the angle between \(AC\) and the base \(BCDE\).
Give your answer correct to 3 significant figures.
(6)
| Scheme | Marks |
|---|---|
| \(360 = (10 \times 10) + 4 \times 0.5 \times 10 \times\) “\(h\)” oe | M1 |
| \(h = 13\) | A1 |
\(AC = \sqrt{13^2 + 5^2}\) = (13.93 or \(\sqrt{194}\)) or \(AO = \sqrt{13^2 - 5^2}\) = (12) or \(OC = (\sqrt{10^2 + 10^2}) \div 2\) = (7.07 or \(5\sqrt{2}\)) or \(EC\) (oe) = \(\sqrt{10^2 + 10^2}\) = (14.14 or \(10\sqrt{2}\)) | M2 |
\(\tan^{-1}\left(\dfrac{12}{7.07}\right)\) or \(\cos^{-1}\left(\dfrac{7.07}{13.93}\right)\) or \(\sin^{-1}\left(\dfrac{12}{13.93}\right)\) or \(\cos^{-1}\left(\dfrac{13.93^2 + 7.07^2 - 12^2}{2 \times 13.93 \times 7.07}\right)\) or \(\cos^{-1}\left(\dfrac{13.93^2 + 14.14^2 - 13.93^2}{2 \times 13.93 \times 14.14}\right)\) | M1 |
| 59.5° | A1 |
| (6) | |
| (6 marks) |
Notes
M1: Finding the perpendicular height of a triangular face
M2: Finding the accurate length of two sides relevant to finding correct angle.
M2 for two sides found or M1 for one side.
1dp rounded or truncated.
M1: A correct trigonometric expression to find correct angle
Accept \(\tan\theta = \left(\dfrac{12}{7.0}\right)\) etc
A1: Accept 59.4° – 59.7°