What is the area of the segment formed by a chord in a circle of radius 8 cm, if the angle subtended at the center is 60°?
\(\dfrac{32\pi}{3} - 16\sqrt{3}\)
Segment area = Sector area − Triangle area (the triangle formed by the two radii and the chord).
Step 1 – Sector area for r = 8 and θ = 60°:
\(A_{sector} = \dfrac{\theta}{360^\circ} \times \pi r^2 = \dfrac{60}{360} \times \pi \times 64 = \dfrac{64\pi}{6} = \dfrac{32\pi}{3}\)
Step 2 – Triangle area (two radii = 8 each, included angle 60°):
\(A_{tri} = \tfrac{1}{2} r^2 \sin\theta = \tfrac{1}{2} \times 64 \times \sin 60^\circ = 32 \times \tfrac{\sqrt{3}}{2} = 16\sqrt{3}\)
Step 3 – Segment area:
\(A_{segment} = \dfrac{32\pi}{3} - 16\sqrt{3}\)
Hence the answer is option 1.
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