Higher November 2021 Paper 1 Q27
27 The vertices of a regular hexagon lie on a circle with centre \(O\) and radius 5 cm
Not drawn accurately
Work out the shaded area.
Give your answer in the form \(\quad \dfrac{a\pi - b\sqrt{c}}{12} \quad\) where \(a\), \(b\) and \(c\) are integers. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(5^2 \times \pi\) (\(\div 6\)) or \(25\pi\) (\(\div 6\)) | M1 | oe allow 3.14 or better for \(\pi\) throughout |
| \(\dfrac{1}{2} \times 5 \times 5 \times \sin 60\) or \(\dfrac{1}{2} \times 5 \times 2.5 \tan 60\) or \(\dfrac{25}{2} \times \dfrac{\sqrt{3}}{2}\) | M1 | oe correct method to work out the area of the triangle or the area of the hexagon implied by \(75 \sin 60\) or \(37.5 \tan 60\) or \(\dfrac{75\sqrt{3}}{2}\) oe |
| \(\dfrac{25\pi}{6} - \dfrac{25\sqrt{3}}{4}\) | A1 | oe eg \(\dfrac{1}{6}\left(25\pi - \dfrac{75\sqrt{3}}{2}\right)\) implied by correct answer |
| \(\dfrac{50\pi - 75\sqrt{3}}{12}\) | A1 | oe in correct form eg \(\dfrac{50\pi - 15\sqrt{75}}{12}\) |
Additional guidance
| Using Pythagoras to work out the perpendicular height of the triangle may lead to an area of \(\dfrac{5\sqrt{18.75}}{2}\) for the triangle or \(15\sqrt{18.75}\) for the area of the hexagon | 2nd M1 |