Date | May 2008 | Marks available | 5 | Reference code | 08M.1.hl.TZ1.3 |
Level | HL only | Paper | 1 | Time zone | TZ1 |
Command term | Find | Question number | 3 | Adapted from | N/A |
Question
A circular disc is cut into twelve sectors whose areas are in an arithmetic sequence.
The angle of the largest sector is twice the angle of the smallest sector.
Find the size of the angle of the smallest sector.
Markscheme
METHOD 1
If the areas are in arithmetic sequence, then so are the angles. (M1)
⇒Sn=n2(a+l)⇒122(θ+2θ)=18θ M1A1
⇒18θ=2π (A1)
θ=π9 (accept 20∘) A1
[5 marks]
METHOD 2
a12=2a1 (M1)
122(a1+2a1)=πr2 M1A1
3a1=πr26
32r2θ=πr26 (A1)
θ=2π18=π9 (accept 20∘) A1
[5 marks]
METHOD 3
Let smallest angle = a , common difference = d
a+11d=2a (M1)
a=11d A1
Sn=122(2a+11d)=2π M1
6(2a+a)=2π (A1)
18a=2π
a=π9 (accept 20∘) A1
[5 marks]
Examiners report
Stronger candidates had little problem with this question, but a significant minority of weaker candidates were unable to access the question or worked with area and very quickly became confused. Candidates who realised that the area of each sector was proportional to the angle usually gained the correct answer.