Higher June 2019 Paper 2 Q19
19 \(ABCD\) is a quadrilateral.
Diagram NOT accurately drawn
The area of triangle \(ACD\) is 250 cm2
Calculate the area of the quadrilateral \(ABCD\).
Show your working clearly.
Give your answer correct to 3 significant figures.
(6)
| Scheme | Marks |
|---|---|
| 250 = 0.5 × 26 × \(AC\) × sin(39) oe | M1 |
| (\(AC\)=) 30.5(5579…) or 30.6 | A1 |
\(\dfrac{(AB)}{\sin 47} = \dfrac{\text{``}{30.56}\text{''}}{\sin 95}\) oe or \(\dfrac{(BC)}{\sin(180 - 95 - 47)} = \dfrac{\text{``}{30.56}\text{''}}{\sin 95}\) oe | M1 |
\((AB =)\;\dfrac{\text{``}{30.56}\text{''}}{\sin 95} \times \sin 47\) (= 22.4(3407…)) or \((BC =)\;\dfrac{\text{``}{30.56}\text{''}}{\sin 95} \times \sin(180 - 95 - 47)\) (= 18.8(8524…)) | M1 |
| 250 + 0.5 × ‘30.56’ × ‘22.43’ × sin(180 – 95 – 47) (= 461.03….) or 250 + 0.5 × ‘30.56’ × ‘18.88’ × sin(47) (= 461.03….) | M1 |
| 461 | A1 |
| (6) | |
| (6 marks) |
Notes
M1: for using the area formula correctly
If this mark is awarded then ft on the remaining M marks
M1: dep on M1 for correct substitution into sine rule
M1: (dep on previous M marks) for a correct method to find a missing length or
sight of values in the ranges
22.39 – 22.47 for \(AB\)
18.8 – 18.92 for \(BC\)
M1: for a complete method to find total area
A1: accept 461 - 462