Higher June 2025 Paper 1R Q20
20 The diagram shows cuboid \(ABCDEFGH\)
Diagram NOT accurately drawn
\(AB = 9\) cm \(AF = 7\) cm \(FC = 18\) cm
Calculate the length of \(BC\)
Give your answer correct to 3 significant figures.
(3)
| Scheme | Marks |
|---|---|
eg \((AC^2 =)\;18^2 - 7^2 (= 275)\) or \((AC =)\sqrt{18^2 - 7^2}\left(= \sqrt{275} \text{ or } 5\sqrt{11} \text{ or } 16.5(831\ldots)\right)\) or \((FB^2 =)\;9^2 + 7^2 (= 130)\) or \((FB =)\sqrt{9^2 + 7^2}\left(= \sqrt{130} \text{ or } 11.4(017\ldots)\right)\) or \((GC^2 =)\;18^2 - 9^2 (= 243)\) or \((GC =)\sqrt{18^2 - 9^2}\left(= \sqrt{243} \text{ or } 9\sqrt{3} \text{ or } 15.5(884\ldots)\right)\) or \(18^2 = (BC)^2 + 7^2 + 9^2\) oe | M1 |
eg \(\text{``}{275}\text{''} - 9^2 (= 194)\) or \(\text{``}{16.5\ldots}\text{''}^2 - 9^2 (= 194)\) or \(18^2 - \text{``}{130}\text{''} (= 194)\) or \(18^2 - \text{``}{11.4\ldots}\text{''}^2 (= 194)\) \(\text{``}{243}\text{''} - 7^2 (= 194)\) or \(\text{``}{15.5\ldots}\text{''}^2 - 7^2 (= 194)\) or \(18^2 - 7^2 - 9^2 (= 194)\) or \(\angle FCB = \sin^{-1}\left(\dfrac{\text{``}{11.4}\text{''}}{18}\right)(= 39.3(036\ldots))\) and \(\cos\text{``}{39.3}\text{''} = \dfrac{(BC)}{18}\) or \(\tan\text{``}{39.3}\text{''} = \dfrac{\text{``}{11.4}\text{''}}{(BC)}\) oe | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 13.9 | A1 |
| (3) | |
| (3 marks) |
Notes
M1: for method to find \(AC^2\) or \(AC\) or \(FB^2\) or \(FB\) or \(GC^2\) or \(GC\) or
for a correct equation using \(BC^2\) and 18 and 7 and 9
other longer ways to find \(AC\), \(FB\), \(GC\) may be used but must be a complete method eg
\(\angle FCA = \sin^{-1}\left(\dfrac{7}{18}\right)(= 22.88\ldots)\) and
\(AC = \dfrac{7}{\tan\text{``}{22.88\ldots}\text{''}}\)
M1: for complete method to find \(BC^2\)
other longer ways to find \(BC\) may be used but must be a complete method, leading to a trig equation in \(BC\)
A1: accept 13.8 to 14