A chord of circle subtends an angle 'Theta' 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 8Sin(theta)/2 * Cos(theta) + (PI) = (PI)(Theta)/45


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

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