∆BOC is equilateral, since both OC and OB are radii of the circle with length 4 cm. Then the angle subtended by the minor arc BC has measure 60°. (Note that OA is also a radius.) AB is a diameter of the circle, so the arc AB subtends an angle measuring 180°. This means the minor arc AC measures 120°.

Since ∆BOC is equilateral, its area is √3/4 (4 cm)² = 4√3 cm². The area of the sector containing ∆BOC is 60/360 = 1/6 the total area of the circle, or π/6 (4 cm)² = 8π/3 cm². Then the area of the shaded segment adjacent to ∆BOC is (8π/3 - 4√3) cm².

∆AOC is isosceles, with vertex angle measuring 120°, so the other two angles measure (180° - 120°)/2 = 30°. Using trigonometry, we find

$\sin(30^\circ) = \dfrac{h}{4\,\rm cm} \implies h= 2\,\rm cm$

where $h$ is the length of the altitude originating from vertex O, and so

$\left(\dfrac b2\right)^2 + h^2 = (4\,\mathrm{cm})^2 \implies b = 4\sqrt3 \,\rm cm$

where $b$ is the length of the base AC. Hence the area of ∆AOC is 1/2 (2 cm) (4√3 cm) = 4√3 cm². The area of the sector containing ∆AOC is 120/360 = 1/3 of the total area of the circle, or π/3 (4 cm)² = 16π/3 cm². Then the area of the other shaded segment is (16π/3 - 4√3) cm².

So, the total area of the shaded region is

(8π/3 - 4√3) + (16π/3 - 4√3) = (8π - 8√3) cm²

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