Higher June 2023 Paper 4 Q9
9 The sides of this triangle are given in centimetres.
The perimeter of the triangle is 80 cm.
Not to scale
You must show your working. [5]
Use calculations to show how you decide. [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 16, 30, 34 in any order with correct working | 5 | M1 for \(x + 5 + 3x + 1 + 2x + 8 = 80\) may be implied by a subsequent correct equation M1 for simplifying their equation to \(ax + b = c\) implied by \(6x + 14 = 80\) M1 for the first correct step in solving their \(ax + b = c\) e.g. \(6x = 80 - 14\) M1 for substituting their 11 into \(x + 5\), \(3x + 1\) and \(2x + 8\) Alternative method using trials M2 for at least one complete correct evaluation of \(x + 5 + 3x + 1 + 2x + 8\) If 0 or 1 scored, instead award SC2 for 16, 30, 34 in any order with no or insufficient working If 0 scored SC1 for \(x = 11\) with no or insufficient working | “Correct working” requires evidence of at least M1 leading to \(x = 11\) or M2 if using trials Note for all methods: \(x = 11\) scores M1M1M1 if there is some supporting work but on its own scores SC1 Alternative method M2 for 80 – 5 – 1 – 8 oe or 66 or M1 for 5 + 1 + 8 or 14 M1 for their 66 ÷ their 6 implied by 11 M1 for substituting their 11 into \(x + 5\), \(3x + 1\) and \(2x + 8\) |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Their fully correct conclusion after M2 scored | 3 | M2 for \(\sqrt{(their\ 16)^2 + (their\ 30)^2}\) correctly evaluated or for (their 16)\(^2\) + (their 30)\(^2\) and (their 34)\(^2\) both correctly evaluated OR M1 for \(\sqrt{(their\ 16)^2 + (their\ 30)^2}\) either not evaluated or incorrectly evaluated or for (their 16)\(^2\) + (their 30)\(^2\) and (their 34)\(^2\) either not evaluated or incorrectly evaluated or for (their 16)\(^2\) + (their 30)\(^2\) correctly evaluated | eg 3 marks for Yes, \(\sqrt{16^2 + 30^2} = 34\) Yes, \(16^2 + 30^2 = 1156\) and \(34^2 = 1156\) or eg after 15, 30, 34 No, \(15^2 + 30^2 = 1125\) and \(34^2 = 1156\) Adapt the scheme for equivalent correct methods e.g. Pythagoras using hypotenuse and subtraction |
Appendix: Question 9(b)
Alternative methods
Cosine rule : [cos … =] \(\frac{16^2 + 30^2 - 34^2}{2 \times 16 \times 30}\) = 0 hence angle … = 90
M1 for correct cos rule statement for angle, M1 for 0 OR \(\cos^{-1}\) or arc cos with 90 and A1 for “Yes” (A1 dep on M2)
Algebra :
M1 for both \((x + 5)^2 + (2x + 8)^2 = x^2 + 10x + 25 + 4x^2 + 32x + 64 = 5x^2 + 42x + 89\) and \((3x + 1)^2 = 9x^2 + 6x + 1\) [ leading to \(4x^2 - 36x - 88 = 0\) and \((x - 11)(4x + 8)\) ]
and M1 for substituting \(x = 11\) into \(5x^2 + 42x + 89\) and \(9x^2 + 6x + 1\) and getting 1156 or for accepting \(x = 11\) and rejecting \(x = -2\)
for correct reason “Yes”
Trigonometry(not cos rule) :
The use of sin, cos or tan to find the other two acute angles will not lead to an exact angle of 90° therefore award just M1 for two correct statements, one about each angle e.g \(\sin[\ldots] = \frac{16}{34}\) and \(\sin[\ldots] = \frac{30}{34}\) and M1 for two angles to at least 2 dp e.g. 28.07[2…] and 61.92[7…] or 61.93 an award max. of M2 because this method will not show exactly that it is a right-angle. The one exception is \(\sin^{-1}\left(\frac{16}{34}\right) + \sin^{-1}\left(\frac{30}{34}\right)\) etc which, if seen, could score 3 marks.