Higher June 2018 Paper 2 Q23
23. Mark’s little sister Lucy has lost a piece of her jigsaw puzzle.
Mark has recorded some of the measurements of the gap left in the jigsaw by this missing piece.
Diagram not drawn to scale
Mark agrees to make a replacement jigsaw piece for Lucy.
One face of the replacement jigsaw piece is to be painted gold.
It cost Mark £3.59 to buy a small pot of gold paint.
The label on the pot states there is enough paint in the pot to cover an area of 60 cm2.
He says Lucy has to pay for the share of the gold paint he uses to make the missing jigsaw piece.
- Calculate the size of \(B\hat{C}D\).
- Hence calculate how much Mark should charge Lucy.
You must show all your working. [10]
| Scheme | Marks | Comments |
|---|---|---|
| \(BD^2 = 4.2^2 + 3.9^2 - 2\times4.2\times3.9\times\cos86^\circ\) | M1 | |
| \(BD^2 = 30.564\ldots.\) or \(BD = 5.528\ldots.\)(cm) | A1 | Accept rounded or truncated, or implied in the next stage of working |
| \(\cos C = \dfrac{6.4^2 + 5.8^2 - BD^2}{2\times6.4\times5.8}\) \((=0.59\ldots)\) | M2 | With either the value for \(BD\) used or the values substituted into the cosine rule from the left hand side triangle M1 for \(BD^2 = 6.4^2 + 5.8^2 - 2\times6.4\times5.8\times\cos C\) |
| An answer in the range 53(°) to 54(°) | A1 | Must be from correct working shown |
| Area \(ABD = \tfrac12\times4.2\times3.9\times\sin86^\circ\) Area \(BCD = \tfrac12\times6.4\times5.8\times\sin C\) | M1 M1 | A value for \(C\) must be shown, FT ‘their derived \(B\hat{C}D\)’ |
| Area \(ABD = 8.17\)(cm2) or 8.2(cm2) OR Area \(BCD = 14.8\) to 15.02 (cm2) | A1 | Only accept 8(cm2) from sight of correct working FT ‘their derived \(B\hat{C}D\)’ |
| \(\left(\dfrac{\text{Total area}}{60}\times3.59 =\right)\) \(\dfrac{22.97}{60}\times3.59\) to \(\dfrac{23.2(0\ldots)}{60}\times3.59\) | M1 | FT ‘their total area’ provided at least 2 marks previously awarded in working to find the area |
| Answer in the range (£)1.37 to (£)1.39 | A1 | CAO. Must follow correct working shown |
| (10) |
Notes
Maximum of SC7 for incorrectly considering \(AC\) as bisecting \(B\hat{A}D\):
| Correctly finding \(BCA\) (sine rule) | B1 |
| Finding \(CBA\) (angle sum triangle) | B1 FT |
| Correctly finding \(AC\) (cosine rule) | B1 FT |
| Correctly finding \(CDA\) (cosine rule) | B1 FT |
| Correctly finding the area of each of the triangles | B1 B1 FT |
| Correctly costs AND finds the charge | B1 |
Corrected by Westie's Workshop
- The board's mark scheme printed “bisecting \(B\hat{A}C\)”. Changed to “bisecting \(B\hat{A}D\)” because line AC is one of the sides of angle BAC, so it cannot cut that angle in half; the common mistake is to assume AC cuts angle BAD in half.
- The board's mark scheme printed “finds the change”. Changed to “finds the charge” because the question asks how much Mark should charge Lucy; there is no change to work out.
Original: see the board's mark scheme PDF (link above).