Higher June 2019 Paper 2 Q16
16 \(ABC\) and \(ACD\) are triangles.
Not drawn accurately
The area of \(ACD\) is 80.5 cm2
Work out the area of \(ABC\).
Give your answer to 3 significant figures. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\dfrac{1}{2} \times 14 \times AC = 80.5\) | M1 | oe eg \(7AC = 80.5\) any letter for \(AC\) |
| \(\dfrac{80.5 \times 2}{14}\) or \(\dfrac{161}{14}\) or 11.5 | M1dep | oe eg \(\dfrac{80.5}{7}\) implies M2 may be seen on diagram |
| \(\dfrac{1}{2} \times 19 \times\) their \(11.5 \times \sin 36\) or 64.21... or 64.22 or 64 | M1 | oe 64.21... or 64.22 or 64 scores M3 if no incorrect formula used |
| 64.2 with no incorrect formula used | A1 |
Additional guidance
| Answer 64.2 with no incorrect working | M3A1 |
| 11.5 scores M2 even if not subsequently used | |
| Answer 64.2 from using ‘\(bh\)’ and ‘\(ab\sin C\)’ (unless clear explanation that \(\dfrac{1}{2}\) has been cancelled in both area formulae) \(14 \times AC = 80.5\) \(\dfrac{80.5}{14} = 5.75\) \(19 \times 5.75 \times \sin 36\) 64.2 | M0 M0 M0 A0 |
| 3rd M1 can be scored if they have a value for \(AC\) eg \(AC = 6\) (may be seen on diagram) \(\dfrac{1}{2} \times 19 \times 6 \times \sin 36 = 33.5\) | M0M0 M1A0 |
| 3rd M1 may be seen in stages eg1 \(11.5 \times \sin 36\) or [6.7, 6.8] \(\dfrac{1}{2} \times 19 \times\) [6.7, 6.8] eg2 19 sin 36 or [11.1, 11.2] \(\dfrac{11.5 \times [11.1, 11.2]}{2}\) |