Higher January 2023 Paper 1R Q17
17
Diagram NOT accurately drawn
\(AB = 4.6\) cm
\(BC = 8.3\) cm
angle \(ABC\) is acute
The area of triangle \(ABC\) is 12 cm²
Work out the perimeter of triangle \(ABC\)
Give your answer correct to 3 significant figures.
(5)
| Scheme | Marks |
|---|---|
| \(12 = \dfrac{1}{2} \times 4.6 \times 8.3 \times \sin ABC\) or \(\dfrac{4.6h}{2} = 12\;(h = 5.217\ldots)\) | M1 |
\(ABC = \sin^{-1}\left(\dfrac{12}{\frac{1}{2} \times 4.6 \times 8.3}\right)\;(= 38.947\ldots)\) oe or \(ABC = \sin^{-1}(0.6286)\;(= 38.947\ldots)\) or \(ABC = \sin^{-1}\left(\dfrac{\text{``}{5.217\ldots}\text{''}}{8.3}\right)\;(= 38.947\ldots)\) or \(BM^2 = 8.3^2 - \text{``}{5.217\ldots}\text{''}^2\) | M1 |
\(AC^2 = 4.6^2 + 8.3^2 - 2 \times 4.6 \times 8.3 \times \cos(\text{``}{38.947}\text{''})\) [allow cos 39°] or \(AC^2 = 30.6(627\ldots)\) \(BM = \sqrt{8.3^2 - \text{``}{5.217\ldots}\text{''}^2}\;(= 6.455\ldots)\) | M1 |
or \(AC = \sqrt{\text{``}{30.6(6\ldots)}\text{''}}\) or 5.5(3739…) | A1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 18.4 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: a correct equation for the area to find angle \(ABC\) or to find the perpendicular height of the triangle.
M1: A correct method to find angle \(ABC\)
or
a correct method to find \(BM^2\) where \(CMB\) is 90°
M1: a correct start to the cosine rule to find length \(AC\)
or a fully correct method for \(BM\)
A1: A correct value for \(AC\) which can be the square root of 30.6(6…)
A1: Allow answers in range 18.4 to 18.45