Higher June 2024 Paper 1R Q15
15 The diagram shows isosceles triangle \(EFG\)
Diagram NOT accurately drawn
\(EF = GF\)
Angle \(EFG = 130^\circ\)
The area of triangle \(EFG\) is 74 cm²
Work out the length of \(EF\)
Give your answer correct to 3 significant figures.
(3)
| Scheme | Marks |
|---|---|
| eg 0.5 × \(EF\) × \(FG\) × sin130 = 74 oe or eg \(EF\) × \(FG\) × sin130 = 2 × 74 oe | M1 |
| \((EF^2 =)\, \dfrac{2 \times 74}{\sin 130}\) (= 193.2...) oe or \((EF =)\sqrt{\dfrac{2 \times 74}{\sin 130}}\;\left(= \sqrt{193.2...}\right)\) oe | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 13.9 | A1 |
| (3) | |
| (3 marks) |
Notes
M1: for setting up an equation using the area of a triangle formula
M1: for a complete method to find \(EF^2\) or \(EF\)
A1: awrt 13.9