Consider a circle with centre O and radius ‘r’ Mark two points A and B on the circle so that
Join AB. Draw BM ⊥ OA
From the figure it is clear that.
Area of ∆OAB < Area of sector OAB < sector of ∆OAC …. (1)
Area of ∆OAB = 1/2 OA.BM
[In ∆OBM. Sin θ BM/r ⇒ BM = r Sinθ]