Higher January 2021 Paper 2 Q20
20
Diagram NOT accurately drawn
The diagram shows a shaded region T formed by removing an equilateral triangle \(PQR\) from a regular hexagon \(ABCDEF\).
The points \(P\) and \(Q\) lie on \(AB\) such that \(AB = 1.5 \times PQ\)
Given that the area of region T is \(72\sqrt{3}\) cm2
work out the length of \(PQ\).
(4)
| Scheme | Marks |
|---|---|
eg \(0.5 \times x \times x \times \sin 60\;\left(= \dfrac{\sqrt{3}}{4}x^2 = 0.433...x^2\right)\) oe where \(x = PQ\) eg \(0.5 \times 2n \times 2n \times \sin 60\;\left(= \sqrt{3}n^2 = 1.732...n^2\right)\) oe where \(2n = PQ\) or use \(0.5 \times b \times h\) where \(h = \sqrt{x^2 - (0.5x)^2}\;\left(= \dfrac{\sqrt{3}}{2}x\right)\) oe | M1 |
eg \(6 \times 0.5 \times 1.5x \times 1.5x \times \sin 60\;\left(= \dfrac{27\sqrt{3}}{8}x^2 = 5.845...x^2\right)\) oe eg \(6 \times 0.5 \times 3n \times 3n \times \sin 60\;\left(= \dfrac{27\sqrt{3}}{2}n^2 = 23.382...n^2\right)\) oe or eg \(2\left(\dfrac{1}{2} \times 1.5x \times 1.5x \times \sin 120\right) + 1.5x \times AE\) where \(AE = \sqrt{(1.5x)^2 + (1.5x)^2 - 2 \times 1.5x \times 1.5x \times \cos 120}\) \(\left(= \dfrac{27\sqrt{3}}{8}x^2 = 5.845...x^2\right)\) or use of \(6 \times 0.5 \times b \times h\), finding \(h\) by Pythagoras | M1 |
eg \(6 \times 0.5 \times 1.5x \times 1.5x \times \sin 60 - 0.5 \times x \times x \times \sin 60 = 72\sqrt{3}\) oe or \(\left(\dfrac{27\sqrt{3}}{8} - \dfrac{\sqrt{3}}{4}\right)x^2 = 72\sqrt{3}\) or (5.845… – 0.433…)\(x^2\) = 124.7… or eg \(6 \times 0.5 \times 3n \times 3n \times \sin 60 - 0.5 \times 2n \times 2n \times \sin 60 = 72\sqrt{3}\) oe \(\left(\dfrac{27\sqrt{3}}{2} - \sqrt{3}\right)n^2 = 72\sqrt{3}\) or (23.382… – 1.732…)\(n^2\) = 124.7… | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 4.8 | A1 |
| (4) | |
| (4 marks) |
Notes
M1: For expression for area of triangle
[using \(AB = x\) and \(PQ = \dfrac{2}{3}x\) gives \(\dfrac{\sqrt{3}}{9}x^2 = 0.192...x^2\)] (correct expression in 1 variable eg \(PQ\))
M1: for expression for area of hexagon
[using \(AB = x\) and \(PQ = \dfrac{2}{3}x\) gives \(\dfrac{3\sqrt{3}}{2}x^2 = 2.598...x^2\)]
(correct expression in 1 variable eg \(AB\))
M1: for a correct equation for shaded area
(correct equation in 1 variable, eg \(PQ\) or \(x\) etc)