Higher June 2021 Paper 2 Q10
10 Here is a triangular prism.
Diagram NOT accurately drawn
Work out the volume of the prism.
Give your answer correct to 3 significant figures.
(5)
| Scheme | Marks |
|---|---|
\(11.2^2 - 7.4^2\ (= 70.68)\) or \([x =]\ \cos^{-1}\left(\dfrac{7.4}{11.2}\right)\ (= 48.64\ldots)\) or \([y =]\ \sin^{-1}\left(\dfrac{7.4}{11.2}\right)\ (= 41.35\ldots)\) or \(\sin^{-1}\left(\dfrac{7.4 \sin 90}{11.2}\right)\) | M1 |
eg \(\sqrt{11.2^2 - 7.4^2}\ (= 8.407\ldots)\) or \([h =]\ \sin\text{``}{48.64\ldots}\text{''} \times 11.2\) or \(\tan\text{``}{48.64\ldots}\text{''} \times 7.4\ (= 8.407\ldots)\) or \([h =]\ \cos\text{``}{41.35\ldots}\text{''} \times 11.2\) or \(\dfrac{7.4}{\tan\text{``}{41.35\ldots}\text{''}}\ (= 8.407\ldots)\) | M1 |
| eg \(7.4 \times \text{``}{8.407}\text{''} \div 2\ (= 31.10\ldots)\) or \(7.4 \times \text{``}{8.407}\text{''} \times 15\ (= 933.19\ldots)\) | M1 |
| eg \(\text{``}{31.10}\text{''} \times 15\ (= 466.59\ldots)\) or \(\text{``}{933.19}\text{''} \div 2\ (= 466.59\ldots)\) | M1 |
| Working not required, so correct answer scores full marks (unless from obvious incorrect working) Answer: 467 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: A correct first stage to finding the perpendicular height of the triangular cross section
M1: oe eg \(h = \dfrac{11.2 \sin\text{``}{48.64\ldots}\text{''}}{\sin 90}\)
M1: for method to find area of cross section or volume of cuboid
M1: complete method to find volume of the prism
A1: accept 466 – 467
SCB2 (if M0 awarded) for \(0.5 \times 7.4 \times \sqrt{11.2^2 + 7.4^2} \times 15\ (= 745)\)
or
SCB1 (if M0 awarded) for \(7.4 \times \sqrt{11.2^2 + 7.4^2} \times 15\ (= 1490)\) or \(0.5 \times 7.4 \times \sqrt{11.2^2 + 7.4^2}\ (= 49.6\ldots)\) or \(0.5 \times 7.4 \times 11.2 \times 15\ (= 621.6)\) or 622