Date | May Example questions | Marks available | 2 | Reference code | EXM.3.AHL.TZ0.2 |
Level | Additional Higher Level | Paper | Paper 3 | Time zone | Time zone 0 |
Command term | Show that and Hence | Question number | 2 | Adapted from | N/A |
Question
This question will explore connections between complex numbers and regular polygons.
The diagram below shows a sector of a circle of radius 1, with the angle subtended at the centre being . A perpendicular is drawn from point to intersect the -axis at . The tangent to the circle at intersects the -axis at .
By considering the area of two triangles and the area of the sector show that .
Hence show that .
Let . Working in modulus/argument form find the solutions to this equation.
Represent these solutions on an Argand diagram. Let their positions be denoted by placed in order in an anticlockwise direction round the circle, starting on the positive -axis. Show the positions of and .
Show that the length of the line segment is .
Hence, write down the total length of the perimeter of the regular sided polygon .
Using part (b) find the limit of this perimeter as .
Find the total area of this sided polygon.
Using part (b) find the limit of this area as .
Markscheme
Area triangle A1
Area sector A1
Area triangle A1
So looking at the diagram M1
AG
[5 marks]
Hence and as we have M1R1
AG
[2 marks]
M1A1M1A1
A1A1
A1
A1
[8 marks]
A1
[1 mark]
Bisecting the triangle to form two right angle triangles M1
Length of where M1A1A1
So length is AG
[4 marks]
Length of perimeter is A1
[1 mark]
as M1A1
[2 marks]
Area of so total area is . M1A1A1
[3 marks]
as M1A1
[2 marks]