Consider a circle with unit radius. There are seven adjacent sectors, S1,S2,S3, ..., S7, in the circle such that their total area is 1/8 of the area of the circle. Further, the area of the jth sector is twice that of the (j–1)th sector, for j = 2, ..., 7. What is the angle, in radians, subtended by the arc of S1
at the centre of the circle?

• pi/508 radians
  
  Suppose 'x' is the area of sector S1
  So, total area = x*(1+2+4+8+16+32+64) = 127*x
  
  So, area of S1 = pi*1*1/8*127 = pi/127*8
  
  area of S1 = (Angle of S1/(2*pi)) * (pi*1*1)
  So, pi/(127*8) = (Angle of S1)/2
  angle of s1 = 2*pi/127*8 = pi/508 radians
  
• 12 years agoHelpfull: Yes(2) No(0)
• The area of unit circle is pi,bcz the area of circle is pi*r^2 and r=1
  therefore s1+s2+s3......+s7=2*pi/16=pi/8.
  as s2=2s1, so 7 adjacent sectors in form of s1 is
  s1+2s1+4s1+8s1+16s1+32s1+64s1=127s1
  so, area of s1=1/127*1/8=1/1016
  as there are 4 quadrants in circle each of 90 degree so total angle will be 2*pi or 360 degree
  so the angle subtended by the arc of s1 is
  2*pi/1016=pi/508
  
• 10 years agoHelpfull: Yes(0) No(0)