Higher June 2022 Paper 5 Q18
18 The diagram shows a shaded shape made by removing sector OAB from sector OCD.
Both sectors have an angle of 60°.
The radius, OA, of the smaller sector is \(r\) cm.
The ratio of radius OA to radius OC is 2 : 3.
Not to scale
Work out, in terms of \(\pi\) and \(r\), the total length of arc AB and arc CD.
Give your answer in its simplest form.
You must show your working. [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\frac{5\pi r}{6}\) with correct working | 5 | Condone ‘×’ sign oe in simplified answer if otherwise correct e.g. \(\frac{5}{6} \times \pi r\) “correct working” requires M1A1M1A1 Condone \(R\) for \(r\) throughout For method marks, allow use of 3.14, 3.142, 22/7 for \(\pi\) | |
| B4 for correct unsimplified answer with correct working | |||
| OR | |||
| M1 for \(\frac{60}{360} \times [2\times]\ \pi k\) oe | Where \(k\) is numeric or algebraic but does not come from squaring Allow e.g. \(k\) = 2, \(r\), \(d\), 0.4 , 0.4\(x\) | ||
| A1 for \(\frac{60}{360} \times 2\pi r\) oe or better isw incorrect cancelling/simplification | For A1 accept e.g. \(0.333\pi r\) Correct expression implies M1A1 | ||
| AND | |||
| M1 for \(\frac{60}{360} \times [2\times]\ \pi\frac{3k}{2}\) | For M1 must use their previous \(k\) e.g. uses \(k\) = 10 for first M1 then uses 15 here for \(\frac{3k}{2}\) gets 2nd M1 unless the expression is correctly stated as \(\frac{60}{360} \times \pi 3r\) oe which gets M1A1 | ||
| A1 for \(\frac{60}{360} \times \pi 3r\) oe or better isw incorrect cancelling/simplification | Correct expression implies M1A1 | ||
| If 0 or 1 scored, instead award SC2 for final answer \(\frac{5\pi r}{6}\) oe simplified answer with no or insufficient working | |||