Seg AB is a chord of a circle with centre P. If PA = 8 cm and distance of chord AB from the centre P is 4 cm, find the area of the shaded portion (minor segment AB). (π = 3.14, √3 = 1.73)

Answer: ≈ 39.31 cm²

Let PM ⊥ AB, PM = 4, PA = 8. In right ΔPMA, \(\cos\angle APM=\dfrac48=\dfrac12\), so \(\angle APM=60^\circ\) and \(\angle APB=120^\circ\).

\(AM=\sqrt{64-16}=4\sqrt3\), so \(AB=8\sqrt3\).

Area of segment = area of sector − area of triangle formed by the two radii and the chord.

\[A(\text{sector})=\frac{120}{360}\times3.14\times64=66.99\]

\[A(\Delta APB)=\tfrac12\times8\sqrt3\times4=16\sqrt3=27.68\]

\[A(\text{shaded})=66.99-27.68=39.31\ \text{cm}^2\]

(The textbook key gives 39.28 sq.cm; the small difference is from rounding.)