Question

Achord of a circle subtend an angle of at the center of circle. The area of the minor segment cut off by the chord is one eighth of the area of the circle . prove that

Answer 1

The question is incomplete dear, but my best guess it is:

A chord of circle subtends an angle θ at centre of the circle. 
The area of the minor segment cut off by the chord is one eighth of the area of the circle. 
Prove that 8(sinθ/2).(cosθ/2) + π = (πθ)/45

This assumption comes as the most asked on from your words. Please see to that uh take care from next time as it is wrong at brainly.in , hope uh get me :)

Anyways, here's your answer:

given : chord AB subtend angle θ at center O.
let the radius of the circle be r.
area of minor segment cut off by the chord AB 
= area of sector AOB - area of triangle OAB
= π r 2 × θ 360 - 1 2 r 2 sin θ 
Since area of this minor segment = 1/8 × area of circle
  πr 2 × θ 360 - 1 2 r 2 sinθ = 1 8 × π r 2 
π θ 360 - sin θ 2 = π 8 
π θ 45 - 4 sin θ = π 1 
π + 4 sin θ = π θ 45 
π + 4 × 2 sin(θ/2)cos(θ/2) = π θ 45 
π + 8 sin(θ/2)cos(θ/2) = π θ 45

Refer to the first image for this method. There is one other method too for solving this, I have attached it ;)

Hope This Helps :)