Higher June 2018 Paper 1 Q9
9 The diagram shows a right-angled triangle.
Diagram NOT accurately drawn
Five of these triangles are put together to make a shape.
Diagram NOT accurately drawn
Calculate the perimeter of the shape.
Give your answer correct to 3 significant figures.
(5)
| Scheme | Marks |
|---|---|
E.g. \(\tan 72 = \dfrac{12.8}{a}\) or \(\tan(90 - 72) = \dfrac{o}{12.8}\) or \(\sin 72 = \dfrac{12.8}{h}\) or \(\cos(90 - 72) = \dfrac{12.8}{h}\) | M1 |
E.g. (shortest side) = \(\dfrac{12.8}{\tan 72}\) or \(12.8\tan(90 - 72)\) or 4.15(89…) or 4.16 or (hypotenuse =) \(\dfrac{12.8}{\sin 72}\) or \(\dfrac{12.8}{\cos(90 - 72)}\) or 13.4(58…) or 13.5 | M1 |
One of (shortest side =) \(\dfrac{12.8}{\tan 72}\) or \(12.8\tan(90 - 72)\) or 4.15(89…) or 4.16 or \(\sqrt{\text{``}{13.4\ldots}\text{''}^2 - 12.8^2}\) AND One of (hypotenuse =) \(\dfrac{12.8}{\sin 72}\) or \(\dfrac{12.8}{\cos(90 - 72)}\) or 13.4(58…) or 13.5 or \(\sqrt{12.8^2 + \text{``}{4.15\ldots}\text{''}^2}\) | M1 |
| 5 × (“13.4(58…)” – “4.15(89…)”) + 5 × 12.8 or 5 × (“13.4…” + “4.15…” + 12.8) – 10 × “4.15…” | M1 |
| 110 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: substitutes correctly into a trig ratio (including the Sine rule)
M1: for a complete method to find one side of the triangle
M1: for a complete method to find both missing sides of triangle
NB Could use Pythagoras’s theorem with side found – must be a complete correct method
M1: for method to use found lengths to find perimeter
A1: for answer in range 110 - 111