The interior of a circle of radius 2 cm is divided into an infinite number of sectors. The areas of these sectors form a geometric sequence with common ratio k. The angle of the first sector is \(\theta \) radians.

(a)     Show that \(\theta = 2\pi (1 - k)\).

(b)     The perimeter of the third sector is half the perimeter of the first sector.

Find the value of k and of \(\theta \).

(a)     the area of the first sector is \(\frac{1}{2}{2^2}\theta \)     (A1)

the sequence of areas is \(2\theta ,{\text{ }}2k\theta ,{\text{ }}2{k^2}\theta \ldots \)     (A1)

the sum of these areas is \(2\theta (1 + k + {k^2} + \ldots )\)     (M1)

\( = \frac{{2\theta }}{{1 - k}} = 4\pi \)     M1A1

hence \(\theta = 2\pi (1 - k)\)     AG

Note: Accept solutions where candidates deal with angles instead of area.

[5 marks]

(b)     the perimeter of the first sector is \(4 + 2\theta \)     (A1)

the perimeter of the third sector is \(4 + 2{k^2}\theta \)     (A1)

the given condition is \(4 + 2{k^2}\theta = 2 + \theta \)     M1

which simplifies to \(2 = \theta (1 - 2{k^2})\)     A1

eliminating \(\theta \), obtain cubic in k: \(\pi (1 - k)(1 - 2{k^2}) - 1 = 0\)     A1

or equivalent

solve for k = 0.456 and then \(\theta = 3.42\)     A1A1

[7 marks]

Total [12 marks]

This was a disappointingly answered question.

Part(a) - Many candidates correctly assumed that the areas of the sectors were proportional to their angles, but did not actually state that fact.

Part(b) - Few candidates seem to know what the term ‘perimeter’ means.