Higher November 2023 Paper 5 Q23
23 In the diagram,
- ABC is a right-angled triangle
- ACD is the sector of a circle with centre A.
Not to scale
(a) Show that the area of the sector ACD is \(\frac{8}{3}\pi\) m2. [6]
(b) Work out the total area of the shape ABCD.
Give your answer in the form \(\left(\dfrac{a\sqrt{k}}{b} + \dfrac{8}{3}\pi\right)\) m2. [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| [\(AC =\)] \(\dfrac{4}{\cos 30}\) oe | M2 | M1 for \(\frac{4}{AC} = \cos 30\) oe or for \(\frac{\sin 60}{4} = \frac{\sin 90}{AC}\) oe | Accept any variable for AC provided not incorrect length Accept longer methods using 4 tan 30 to find BC then Pythagoras’ or sine rule M2 for AC explicit |
| (\(AC =\)) \(\frac{8}{\sqrt{3}}\) or better | A2 | B1 for \(\cos 30 = \frac{\sqrt{3}}{2}\) oe | oe for B1 e.g. \(\sin 60 = \frac{\sqrt{3}}{2}\) |
| \(\dfrac{45}{360} \times \pi \times \left(\textit{their}\ \dfrac{8}{\sqrt{3}}\right)^2\) | M1 | dep on at least M1 | |
| \(\frac{45}{360} \times \pi \times \frac{64}{3}\) or better or \(\frac{1}{8} \times \pi \times \frac{8}{\sqrt{3}} \times \frac{8}{\sqrt{3}}\) \(= \frac{8}{3}\pi\) | A1 | with no errors seen | Accept \(\frac{1}{8} \times \pi \times \frac{8^2}{3}\) |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\dfrac{8\sqrt{3}}{3} + \dfrac{8}{3}\pi\) | 3 | M1 for \(\frac{1}{2} \times 4 \times \frac{8}{\sqrt{3}} \times \sin 30\) FT their AC from (a) or for \(\frac{1}{2} \times 4 \times 4 \times \tan 30\) B1 for \(\sin 30 = \frac{1}{2}\) oe or for \(\tan 30 = \frac{\sqrt{3}}{3}\) oe | Accept e.g. \(\frac{16\sqrt{3}}{6} + \frac{8}{3}\pi\) for 3 marks |