Higher November 2024 Paper 2 Q25
25 The diagram shows an equilateral triangle \(ABC\) and a circle with centre \(O\)
Diagram NOT accurately drawn
\(AB\), \(BC\) and \(CA\) are tangents to the circle.
The radius of the circle is \(x\) cm
The total area, in cm², of the regions shown shaded in the diagram is \(nx^2\)
Find the value of \(n\)
Give your answer correct to 3 significant figures.
(5)
| Scheme | Marks |
|---|---|
eg \(\tan 30 = \dfrac{x}{0.5AB}\) or \(\tan 60 = \dfrac{0.5AB}{x}\) oe or \(0.5AB = \dfrac{x}{\tan 30}\) or \(0.5AB = x\tan 60\) oe or \(\dfrac{1}{2}AB = \sqrt{3}x\) | M1 |
eg \((AB =)\;2x\tan 60\) or \(\dfrac{2x}{\tan 30}\) or \(2\sqrt{3}x\) or \(3.46..x\) oe OR area small triangle \(\left(\dfrac{1}{6}\textit{shape}\right) = \dfrac{1}{2} \times x\tan 60 \times x\) or \(\dfrac{1}{2} \times \sqrt{3}x \times x\) or \(\dfrac{\sqrt{3}}{2}x^2\) | M1 |
eg \(6 \times \dfrac{1}{2} \times x\tan 60 \times x\) or \(\dfrac{1}{2} \times (2x\tan 60)^2 \times \sin 60\) or \(\dfrac{1}{2}\left(\dfrac{2x}{\tan 30}\right)^2 \sin 60\) or \(\dfrac{1}{2} \times 2\sqrt{3}x \times \sqrt{\left(2\sqrt{3}x\right)^2 - \left(\sqrt{3}x\right)^2}\) or \(0.5 \times 2\sqrt{3}x \times 3x\) or \(3\sqrt{3}x^2\) or \(5.19\ldots x^2\) | M1 |
\(6 \times \dfrac{\sqrt{3}}{2}x^2 - \pi x^2\) or \(3\sqrt{3}x^2 - \pi x^2\) or \(\text{``}{5.19\ldots}\text{''}x^2 - \pi x^2\) or \(6 \times \dfrac{1}{2} \times \tan 60 \times x^2 - \pi x^2\) or \(\dfrac{1}{2}\left(\dfrac{2x}{\tan 30}\right)^2 \sin 60 - \pi x^2\) oe | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Look for values written on the diagram Answer: 2.05 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: (off spec but a student who knows without calc that height of triangle = \(3x\): \(\tan 60 = \dfrac{3x}{0.5AB}\) or \(\sin 60 = \dfrac{3x}{BC}\) )
M1: Expression for side of triangle (\(AB\) or \(BC\) or \(AC\))
OR
the area of one or more of the six triangles
M1: a correct expression for the area of triangle \(ABC\)
M1: a correct expression for the area of the shaded parts of the diagram or a correct equation for the shaded parts of the diagram
A1: 2.05 – 2.06
| Scheme | Marks |
|---|---|
eg \(\tan 30 = \dfrac{100}{0.5AB}\) or \(\tan 60 = \dfrac{0.5AB}{100}\) oe or \(0.5AB = \dfrac{100}{\tan 30}\) or \(0.5AB = 100\tan 60\) oe or \(\dfrac{1}{2}AB = \sqrt{3} \times 100\) | M1 |
eg \((AB =)\;200\tan 60\) or \(\dfrac{200}{\tan 30}\) or \(200\sqrt{3}x\) or \(346\ldots x\) ) or area small triangle \(\left(\dfrac{1}{6}\textit{shape}\right) = \dfrac{1}{2} \times \text{``}{173\ldots}\text{''} \times 100\;(= 8660\ldots)\) | M1 |
eg \(\dfrac{1}{2} \times 200\sqrt{3} \times 200\sqrt{3} \times \sin 60\;(= 51961.52\ldots)\) or or \(6 \times \dfrac{1}{2} \times \text{``}{173\ldots}\text{''} \times 100\;(= 6 \times \text{``}{8660\ldots}\text{''} = 51961.52\ldots)\) | M1 |
| eg \(\text{``}{51961.52}\text{''} - \pi \times 100^2 = 100^2 n\) | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 2.05 | A1 |
Notes
M1: eg using radius = 100
ANY VALUE CAN BE USED CONSISTENTLY FOR AWARD OF MARKS
M1: Expression for side of triangle
or
the area of one or more of the six triangles
M1: a correct expression for the area of the triangle
M1: a correct equation for the area of the shaded parts of the diagram.
A1: 2.05 – 2.06