Higher November 2024 Paper 4 Q16
16 AC is a diagonal of the quadrilateral ABCD.
Not to scale
AD = 72 m.
Angle ABC = 108°, angle BCA = 38° and angle BAC = 34°.
Angle ACD = 90°, angle CDA = 65° and angle CAD = 25°.
Find the area of ABCD.
You must show your working. [6]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 1763 or 1763.4 to 1763.5 with correct working | 6 | “Correct working” requires evidence of at least M2 or M1 M1 condone answers in range 1750 – 1780 with full correct working | |
| M2 for \(\frac{1}{2} \times 72 \times\) their 65.25… × sin 25 or M1 for [AC=] 72cos 25 or 72sin 65 oe | M2 implied by 992.79… or 992.8 or maybe by 985 – 999[…] nfww or M1 implied by [AC=] 65.2[5…] or 65.3 | ||
| M3 for \(\frac{1}{2} \times\) their 65.254 × their 42.24… × sin34 oe or M2 for [AB=] \(\frac{\textit{their } 65.25\ldots \times \sin 38}{\sin 108}\) oe or M1 for \(\frac{\text{AB}}{\sin 38} = \frac{\textit{their } 65.25\ldots}{\sin 108}\) oe | M3 implied by 770.69… or 770.7 or maybe by 760 – 783 nfww M2 implied by [AB=] 42.2[4…] | ||
| If 0, 1 or 2 scored award SC3 for answer 1763 or 1763.4 to 1763.5 with no or insufficient working If 0 or 1 scored award SC2 for 992.79… or 992.8 or 770.69… or 770.7 with no or insufficient working | accept any correct method e.g. M2 for area of triangle ACD = ½ × 72sin25 × 72cos25 oe or M1 for [CD=]72sin25 implied by 30.4… | ||
Appendix: Question 16
Area of ABC = 770.687099… and area of ACD = 992.793598… total area = 1763.480697…
Note : condone answers in range 1745 – 1780, area ACD in range 985 – 998 and area ABC in range 760 – 783