4MA0 Higher June 2015 Paper 3 Q23
23
Diagram NOT accurately drawn
\(ABCD\) is the square base of the pyramid \(VABCD\).
\(AB = BC = CD = DA = 10\) cm.
\(VA = VB = VC = VD = 12\) cm.
Calculate the height of the pyramid.
Give your answer correct to 3 significant figures.
(4)
| Scheme | Marks |
|---|---|
Let \(O\) be the centre of the square. \((AC^2) = 10^2 + 10^2\) (= 200) or \((AC =)\;\sqrt{200}\) oe or \((AC =)\) 14.1(4…) | M1 |
\((AO =)\;\dfrac{1}{2}\sqrt{200}\) oe or \((AO) = 7.07(1\ldots)\) or \((AO) = 7.05\) | M1 |
\((VO^2 =)\;12^2 - \left(\dfrac{1}{2}\sqrt{200}\right)^2\) oe (= 94) OR Angle \(VAC\) is \(\cos^{-1}\left(\dfrac{7.07}{12}\right) = 53.896^\circ\) AND 12 sin 53.896 (= 9.695...) | M1 |
| 9.70 | A1 |
| (4) |
Notes
M1: or \(2AO^2 = 10^2\)
M1: (dep on both previous method marks) for a fully correct method (condone missing brackets)
A1: awrt 9.70 accept 9.7
Alternative method
| Scheme | Marks |
|---|---|
Let \(M\) be the midpoint of a side of the square \(VM^2 = 12^2 - 5^2\) (= 119) or \(VM = \sqrt{119}\) (= 10.9(…)) | M2 |
| \(VO^2 = 119 - 5^2\) (= 94) or \(VO^2 = 10.9^2 - 5^2\) oe | M1 |
| 9.70 | A1 |
Notes
M2: but it must be explicitly clear that it is \(VM\) being calculated
A1: awrt 9.70 accept 9.7