Higher January 2022 Paper 1 Q9
9 The diagram shows isosceles triangle \(ABC\)
Diagram NOT accurately drawn
\(AB = AC = 17.5\) cm \(BC = 28\) cm
Calculate the area of triangle \(ABC\)
(4)
| Scheme | Marks |
|---|---|
| \(17.5^2 - 14^2\) (= 110.25) | M1 |
| \(\sqrt{17.5^2 - 14^2}\) (= 10.5) | M1 |
| 0.5 × 28 × “10.5” oe | M1 |
| 147 | A1 |
| (4) | |
| (4 marks) |
Notes
M1: or for use of cosine rule to find one of the angles eg \(28^2 = 17.5^2 + 17.5^2 - 2 \times 17.5 \times 17.5 \times \cos A\)
or eg \(\cos B = \dfrac{14}{17.5}\)
M1: or for rearranging the cosine rule to eg \(\cos A = \dfrac{17.5^2 + 17.5^2 - 28^2}{2 \times 17.5 \times 17.5}\) (\(A\) = 106.26…)
or eg \(B = \cos^{-1}\left(\dfrac{14}{17.5}\right)\) (= 36.86…)
M1: or for 0.5 × 17.5 × 17.5 × sin106.26… oe
eg 0.5 × 17.5 × 28 × sin36.86…
[clear use of Heron’s formula:
M1 for S = 0.5(17.5 + 17.5 + 28)(=31.5)
M2 for \(\sqrt{\text{``}{31.5}\text{''}(\text{``}{31.5}\text{''} - 17.5)^2(\text{``}{31.5}\text{''} - 28)}\) oe]
A1: accept awrt 147