Higher June 2025 Paper 2 Q24
24 The diagram shows a square-based pyramid \(ABCDE\)
Diagram NOT accurately drawn
\(EA = EB = EC = ED\)
\(M\) is the centre of the horizontal square base \(ABCD\)
\(Q\) is the midpoint of \(AB\)
Angle \(EQM = 80^\circ\)
\(EA : AB = n : 1\)
Find the value of \(n\)
Give your answer correct to 3 significant figures.
(4)
| Scheme | Marks |
|---|---|
To find \(EQ\) eg \(MQ = 0.5\) (= \(AQ\)) \(\dfrac{0.5}{\cos 80}\left(= 2.87(938..)\right)\) eg \(MQ = x\) (= \(AQ\)) \(\dfrac{x}{\cos 80}(= 5.75..x)\) NB cos80 = sin10 or To find \(EM\) eg \(MQ = 0.5\) (= \(AQ\)) \(0.5\tan 80\left(= 2.83(564..)\right)\) eg \(MQ = x\) (= \(AQ\)) \(x\tan 80(= 5.67..x)\) NB \(\tan 80 = \dfrac{1}{\tan 10}\) or To find \(EQ\) eg \(EM = y\) \(\dfrac{y}{\sin 80}(= 1.01..y)\) To find \(MQ\) (= \(AQ\)) eg \(EQ = h\) \(h\cos 80(= 0.173..h)\) | M1 |
To find \(EA^2\) eg \(AQ = 0.5\) (= \(MQ\)) \(\text{``}{2.87..}\text{''}^2 + 0.5^2\) \(\left(= 8.54(08..)\right)\) eg \(AQ = x\) (= \(MQ\)) \(x^2 + \text{``}{\left(\dfrac{x}{\cos 80}\right)}\text{''}^2\) or To find \(AM\) eg \(MQ = 0.5\) (= \(AQ\)) \(\sqrt{0.5^2 + 0.5^2}\) \(\left(= \dfrac{\sqrt{2}}{2} = 0.707(10..)\right)\) eg \(MQ = x\) (= \(AQ\)) \(\sqrt{x^2 + x^2}\left(= \sqrt{2x^2} = x\sqrt{2}\right)\) or To find \(MQ\) (= \(AQ\)) eg \(EM = y\) \(\dfrac{y}{\tan 80}(= 0.176..y)\) To find \(EA^2\) eg \(EQ = h\) \(h^2 + \text{``}{(h\cos 80)}\text{''}^2\) | M1 |
To find \(EA\) eg \(AQ = 0.5\) (= \(MQ\)) \(\sqrt{\text{``}{2.87..}\text{''}^2 + 0.5^2}\) \(\left(= \sqrt{8.54(08..)}\right)\) or eg \(AQ = x\) (= \(MQ\)) \(\sqrt{x^2 + \text{``}{\left(\dfrac{x}{\cos 80}\right)}\text{''}^2}\;(= 5.84..x)\) or To find \(EA\) eg \(AQ = 0.5\) (= \(MQ\)) \(\sqrt{\text{``}{2.83..}\text{''}^2 + \text{``}{0.707..}\text{''}^2}\) \(\left(= \sqrt{8.54(08..)}\right)\) eg \(AQ = x\) (= \(MQ\)) \(\sqrt{\text{``}{(x\tan 80)}\text{''}^2 + \text{``}{(x\sqrt{2})}\text{''}^2}\) \((= 5.84..x)\) or To find \(EA\) eg \(EM = y\) \(\sqrt{\text{``}{\left(\dfrac{1}{\sin 80}\right)}\text{''}^2 + \text{``}{\left(\dfrac{1}{\tan 80}\right)}\text{''}^2}\) \((= 1.03..y)\) [= 1.03.. × \(AQ\) × tan80] eg \(EQ = h\) \(\sqrt{h^2 + \text{``}{(h\cos 80)}\text{''}^2}\;(= 1.01..h)\) \(\left[= 1.01.. \times \dfrac{0.5}{\cos 80}\right]\) | M1 |
| Working not required, so correct answer scores full marks (unless from obvious incorrect working) Answer: 2.92 | A1 |
| (4) | |
| (4 marks) |
Notes
M1: Use of a value for the side \(AB\) eg 1 or \(2x\) or any value etc or Let \(EM = y\) or any value or \(EQ = h\)
A1: awrt 2.92