Question

Consider a circle with unit radius. There are seven adjacent sectors, s1, s2, s3, ..., s7, in the circle such that their total area is 18of 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?

Answer 1

S1 , S2 , S3 , ...........S7 are the adjacent sectors in the circle of radius 1 unit .

let ∅1 , ∅2 , ∅3 .........∅7 are the angle subtend by S1 , S2, S3 .......S7 arc respectively .

area of S1 sector = ∅1/360° × πr²

area of S2 sector = ∅2/360° × πr²

................

.....

area of S7 sector = ∅7/360× πr²

question ask ,

area of j th sector = 2× area of (j -1)th sector .

where j =2, 3, .......7

it means ,

area of S2 sector = 2× area of S1 sector

∅1/360× πr² = 2∅/360 × πr²

∅2 = 2∅1

similarly ,

∅3=2∅2

∅4 =2∅3

∅5 =2∅4

∅6 =2∅5

∅7 =2∅6

if we solve this

we find ,

∅2 = 2∅1

∅3 = 4∅1

∅4 =8∅1

∅5 =16∅1

∅6=32∅1

∅7 =64∅1

again question, ask

area of total sectors = 1/8 area of circle

(∅1 + ∅2 +.....∅7)/360° × πr² = 1/8 ×πr²

(∅1 + ∅2+.....∅7) = 360/8°

so,

(∅1 + 2∅1 +4∅1 +....64∅1) =360°/8 ×2π/360 = π/4

∅1( 1+ 2 +4 + 8 .....64) =π/4

1 + 2 + 4 + 8 + 16....64 are in GP

so,

64 = 1.(2)^(n-1)

n =7

Sn = 1(2^7 -1)/(2-1) = 2^7 -1 = 127

so,

∅1 × 127 =π/4

∅1 = π/127× 4 = π/508

hence , arc S1 subtended π/508 rad

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