Higher June 2024 Paper 3 Q21
21 The diagram shows a circle, radius \(r\) cm and two regular hexagons.
Each side of the larger hexagon \(ABCDEF\) is a tangent to the circle.
Each side of the smaller hexagon \(PQRSTU\) is a chord of the circle.
By considering perimeters, show that
\[3 \lt \pi \lt 2\sqrt{3}\](4)
| Answer | Mark | Mark scheme |
|---|---|---|
| Result shown | P1 | for process to find the length of half a side or a side or the perimeter of the smaller hexagon \(PQRSTU\), eg \(r\sin 30\ \left(= \dfrac{r}{2}\right)\) oe or \(2r\sin 30\ (= r)\) oe or \(6 \times 2r\sin 30\ (= 6r)\) oe |
| P1 | for process to find the length of half a side or a side or the perimeter of the larger hexagon \(ABCDEF\) eg Length of half side \(= r\tan 30\) or \(\dfrac{r}{\tan 60}\ \left(= \dfrac{\sqrt{3}r}{3}\right)\) oe or Length of side \(= 2r\tan 30\) or \(\dfrac{2r}{\tan 60}\) or \(\dfrac{r}{\sin 60}\) or \(\dfrac{r}{\cos 30}\ \left(= \dfrac{2\sqrt{3}r}{3}\right)\) oe or Length of perimeter \(= 6 \times 2r\tan 30\) or \(6 \times \dfrac{2r}{\tan 60}\) oe | |
| P1 | (dep P2) for process of forming a correct inequality, eg using half lengths eg \(\dfrac{r}{2} \lt \dfrac{2\pi r}{12} \lt r\tan 30\) oe or using lengths eg \(r \lt \dfrac{2\pi r}{6} \lt 2r\tan 30\) oe or using perimeters eg \(6 \times r \lt 2\pi r \lt 6 \times 2r\tan 30\) oe | |
| C1 | (dep P2) correct algebra leading to given result, \(3 \lt \pi \lt 2\sqrt{3}\) |
Additional guidance
May use Sine Rule
or cos60 instead of sin30
May use Sine Rule
Note this mark is not for just the sight of \(\dfrac{\sqrt{3}r}{3}\) or \(\dfrac{2\sqrt{3}r}{3}\) or \(\dfrac{12\sqrt{3}r}{3}\) oe, they need to be associated with the correct length
Perimeter \(= 4\sqrt{3}r\) alone does not get this mark