Question

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 radians. (a) show that = 2(1 k). (b) the perimeter of the third sector is half the perimeter of the first sector. Find the value of k and of .

Answer 1

Answer:

1.  θ = 2π(1 – k).

2. θ = 1.85 radians. , k = 1/√2

Step-by-step explanation:

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 θ radians.

(a)Show that θ = 2π(1 – k).

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

Find the value of k and of θ.

Area of sector with θ radians = θr²/2. (r is radius of the circle, θ in radians)

Given all the areas of sectors are in Geometric Progression (GP) with common ratio k.

First term a1 = θr²/2  

Second term a2 = kθr²/2

Third term a3 = k²θr²/2

This is a infinite GP.

The sum of infinite GP = a1/(1 – k)  ( k is common ratio)

Substituting a1, we get  

θr²/2(1 – k)  = π*r²  (The total are of GP is area of the circle)

θ = 2π(1 – k)

2. The perimeter of the third sector is half the perimeter of the first sector.

Perimeter of sector = θr. (r is radius of the circle, θ in radians)

Perimeter of third sector = k²θr

Perimeter of first sector = θr

Given that

K²θr = θr/2

K = 1/√(2)

 

θ = 2π(1 – k) = 2π(1 – 1/√(2)) = π(2 – √(2)) = 1.85 radians.