1MA0 Higher November 2012 Paper 2 Q23
23 The diagram shows a pyramid.
Diagram NOT accurately drawn
\(BCDE\) is a square with sides of length 10 cm.
The other faces of the pyramid are equilateral triangles with sides of length 10 cm.
(a) Calculate the volume of the pyramid.
Give your answer correct to 3 significant figures. (4)
Give your answer correct to 3 significant figures. (4)
(b) Find the size of angle \(DAB\). (2)
| Working | Answer | Mark | Notes |
|---|---|---|---|
| Let \(O\) be the centre of the base. \(OB^2 + OC^2 = 10^2\) ; \(OB^2 = 50\) \(AO^2 = AB^2 - OB^2 = 50\) Vol \(= \frac{1}{3} \times 10^2 \times \sqrt{50}\) OR Let \(M\) be the midpt of side \(BC\) and let \(O\) be the centre of the base. \(AM^2 + MC^2 = 10^2\) ; \(AM^2 = 75\) \(AO^2 = AM^2 - MO^2 = 50\) Vol \(= \frac{1}{3} \times 10^2 \times \sqrt{50}\) | 236 | 4 | M1 correct method to start to find \(BD\) or \(BO\) using triangle \(OBC\) or triangle \(BCD\) (oe) Eg. \(OB^2 + OC^2 = 10^2\) or \(BO^2 = 50\) or \(BO = \sqrt{50}\) (=7.07..) or \(BO = \dfrac{\sqrt{200}}{2}\) or \(10^2 + 10^2 = BD^2\) or \(BD^2 = 200\) or \(BD = \sqrt{200}\) (=14.1..) M1 (dep) correct method to find height of pyramid using triangle \(AOB\) Eg. \(AO^2 = 10^2 -\) ‘\(\sqrt{50}\)’\(^2\) or \(AO^2 = 50\) or \(AO = \sqrt{50}\) (=7.07..) M1 (indep) \(\frac{1}{3} \times 10^2 \times\) ‘\(\sqrt{50}\)’ (but not \(\frac{1}{3} \times 10^2 \times 10\)) A1 235 – 236 OR M1 correct method to start to find height of a face using triangle \(AMC\) (oe) Eg. \(AM^2 + 5^2 = 10^2\) or \(AM^2 = 75\) or \(AM = \sqrt{75}\) (=8.66...) M1 (dep) correct method to find height of pyramid using triangle \(AOM\) Eg. \(AO^2 =\) ‘\(\sqrt{75}\)’\(^2 - 5^2\) or \(AO^2 = 50\) or \(AO = \sqrt{50}\) (=7.07..) M1 (indep) \(\frac{1}{3} \times 10^2 \times\) ‘\(\sqrt{50}\)’ (but not \(\frac{1}{3} \times 10^2 \times 10\)) A1 235 – 236 OR M1 for \(\sin 45 = \dfrac{x}{10}\) or \(\cos 45 = \dfrac{x}{10}\) M1 for \(h = 10 \times \sin 45\) or \(h = 10 \times \cos 45\) (=7.07..) M1 (indep) \(\frac{1}{3} \times 10^2 \times\) ‘7.07...’ (but not \(\frac{1}{3} \times 10^2 \times 10\)) A1 235 – 236 |
| Working | Answer | Mark | Notes |
|---|---|---|---|
| Angle \(ABO = 45^\circ\) Angle \(DAB = 180 - 45 - 45\) OR In \(\Delta BAD\), \(\cos A = \dfrac{10^2 + 10^2 - \text{‘}\sqrt{200}\text{’}^2}{2 \times 10 \times 10} = 0\) OR In \(\Delta BOA\), \(\cos B = \dfrac{\text{‘}\sqrt{50}\text{’}}{10}\) Angle \(BAD = 180 -\) ‘45’ – ‘45’ OR \(\sin A = \dfrac{\text{‘}\sqrt{50}\text{’}}{10}\) \(A = 45\) Angle \(BAD = 2 \times\) ‘45’ | 90 | 2 | M1 Angle \(DAB = 180 - 2 \times\) ‘45’ A1 89.98 - 90 OR M1 \(\cos BAD = \dfrac{10^2 + 10^2 - \text{‘}\sqrt{200}\text{’}^2}{2 \times 10 \times 10}\) A1 89.98 - 90 OR M1 \(\sin A = \dfrac{\text{‘}\sqrt{50}\text{’}}{10}\) A1 89.98 - 90 |