Foundation June 2023 Paper 1R Q20
20 The diagram shows a shape made up of three semicircles, enclosing a right-angled triangle.
Diagram NOT accurately drawn
\(AB\), \(BC\) and \(CA\) are each the diameter of a semicircle.
\(BC = CA = 6\) cm.
Work out the perimeter of the shape.
Give your answer correct to one decimal place.
(5)
| Scheme | Marks |
|---|---|
eg \((AB^2 =)\;6^2 + 6^2\;(= 72)\) or \(\sin 45 = \dfrac{6}{(AB)}\) or \(\cos 45 = \dfrac{6}{(AB)}\) or or \((AB^2 =)\;6^2 + 6^2 - 2 \times 6 \times 6 \times \cos 90\) | M1 |
eg \((AB =)\sqrt{6^2 + 6^2}\;\left(= \sqrt{72} \text{ or } 6\sqrt{2} \text{ or } 8.48...\right)\) or \((AB =)\dfrac{6}{\sin 45}\left(= \sqrt{72} = 6\sqrt{2} = 8.48...\right)\) or \((AB =)\dfrac{6}{\cos 45}\left(= \sqrt{72} = 6\sqrt{2} = 8.48...\right)\) or \((AB =)\sqrt{6^2 + 6^2 - 2 \times 6 \times 6 \times \cos 90}\) | M1 |
| eg \(\pi \times 6\;(= 6\pi \text{ or } 18.8...)\) or \(\pi \times 6 \div 2\;(= 3\pi \text{ or } 9.42...)\) or \(\pi \times \text{“}{8.48...}\text{”}\;(= 26.6...)\) or \(\pi \times \text{“}{8.48...}\text{”} \div 2\;(= 13.3...)\) | M1 |
| eg \(2 \times \text{“}{3\pi}\text{”} + \text{“}{13.3...}\text{”}\) or “9.42” + “9.42” + “13.3” or “18.8” + “13.3” | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 32.2 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for a correct start to the method to find \(AB\)
M1: for a complete method to find the length of \(AB\)
M1: (indep) for a method to find the circumference of one whole circle or the arc length of one semicircle seen (may be embedded)
M1: for a complete correct method to find the perimeter of the shape
A1: accept answers in the range 32.1 – 32.3