Foundation June 2021 Paper 2 Q24
24 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...) or [\(y\) =] \(\sin^{-1}\left(\dfrac{7.4}{11.2}\right)\) (= 41.35...) or \(\sin^{-1}\left(\dfrac{7.4 \sin 90}{11.2}\right)\) | M1 |
eg \(\sqrt{11.2^2 - 7.4^2}\) (= 8.407...) or [\(h\) =] \(\sin \text{“}48.64\ldots\text{”} \times 11.2\) or \(\tan \text{“}48.64\ldots\text{”} \times 7.4\) (= 8.407...) or [\(h\) =] \(\cos \text{“}41.35\ldots\text{”} \times 11.2\) or \(\dfrac{7.4}{\tan \text{“}41.35\ldots\text{”}}\) (= 8.407…) | M1 |
| eg 7.4 × “8.407” ÷ 2 (= 31.10....) or 7.4 × “8.407” × 15 (= 933.19...) | M1 |
| eg “31.10” × 15 (= 466.59...) or “933.19” ÷ 2 (= 466.59...) | 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…) or
\(0.5 \times 7.4 \times 11.2 \times 15\) (= 621.6) or 622