Higher June 2022 Paper 1R Q18
18 The diagram shows a cube \(ABCDEFGH\) with sides of length 6 cm.
Diagram NOT accurately drawn
\(T\) is the midpoint of \(AB\) and \(V\) is the midpoint of \(CH\)
Work out the distance from \(T\) to \(V\) in a straight line through the cube.
Give your answer in the form \(\sqrt{a}\) cm where \(a\) is an integer.
(4)
| Scheme | Marks |
|---|---|
| eg \((BV^2 =)\;3^2 + 6^2\) (= 45) or \((CT^2 =)\;3^2 + 6^2\) (= 45) or \((DH^2 =)\;6^2 + 6^2\) (= 72) or \((MV^2 =)\;3^2 + 3^2\) (= 18) | M1 |
eg \((BV =)\;\sqrt{3^2 + 6^2}\;\left(= \sqrt{45}\text{ or }3\sqrt{5}\text{ or }6.70\ldots\right)\) or \((CT =)\;\sqrt{3^2 + 6^2}\;\left(= \sqrt{45}\text{ or }3\sqrt{5}\text{ or }6.70\ldots\right)\) or \((DH =)\;\sqrt{6^2 + 6^2}\;\left(= \sqrt{72}\text{ or }6\sqrt{2}\text{ or }8.48\ldots\right)\) or \((MV =)\;\sqrt{3^2 + 3^2}\;\left(= \sqrt{18}\text{ or }3\sqrt{2}\text{ or }4.24\ldots\right)\) | M1 |
\((VT =)\;\sqrt{\text{``}{45}\text{''} + 3^2}\) or \(\sqrt{\left(\dfrac{\text{``}{\sqrt{72}}\text{''}}{2}\right)^2 + 6^2}\) or \(\sqrt{\text{``}{18}\text{''} + 6^2}\) or \(3\sqrt{6}\) or 7.34… | M1 |
| \(\sqrt{54}\) | A1 |
| (4) | |
| (4 marks) |
Notes
M1: a correct expression for eg \(BV^2\) or \(CT^2\) or \(DH^2\) or \(MV^2\) where \(M\) is the midpoint of \(DC\) or a correct expression for [length]2 for any length in the cube using Pythagoras
M1: for a complete method for eg \(BV\) or \(CT\) or \(DH\) or \(MV\) or any length in the cube using Pythagoras
M1: for a correct expression for \(VT\) (condone missing brackets around \(3\sqrt{5}\) or \(3\sqrt{2}\) or \(\dfrac{\sqrt{72}}{2}\))
A1: if \(\sqrt{54}\) seen and answer then given as \(3\sqrt{6}\) isw and award full marks
M3 for \((VT =)\;\sqrt{6^2 + 3^2 + 3^2}\) (\(= 3\sqrt{6}\) or 7.34…)
(M2 for \((VT^2 =)\;6^2 + 3^2 + 3^2\) (= 54))