The radius of a circle is R.AB is a chord of this circle. AB cuts the circle into 2 unequal segments. C is a point on the smaller arc AB. If the angle ACB is equal to 120º, then the area of the smaller segment is
$\frac{{{R}^{2}}}{12}\left( 5\pi -3\sqrt{3} \right)$
$\frac{{{R}^{2}}}{16}\left( 4\sqrt{2}-2\pi \right)$
$\frac{{{R}^{2}}}{8}\left( 8\pi -10\sqrt{3} \right)$
$\frac{{{R}^{2}}}{12}\left( 5\pi -4\sqrt{3} \right)$
Solution
ABCD is a quadrilateral.
$\angle ACB = 120^{\circ}$ then,
$\angle ADB = 60^{\circ}$
$\angle AOB = 120^{\circ}$
Area of segment AOBCA
$\Rightarrow \dfrac{120^{\circ}}{360} \times \pi \times r^2 = \dfrac{\pi}{3}r^2$ (r = radius of circle)
Area of $\Delta AOB$
$= \dfrac{1}{2} \times r \times r \times \sin 120^{\circ} \Rightarrow \dfrac{r^2}{2} \times \dfrac{\sqrt{3}}{2} = \dfrac{\sqrt{3}r^2}{4}$
Area of segment ABC = Area of segment AOBCA - Area of $\Delta AOB$
$= \dfrac{\pi}{3}r^2 - \dfrac{\sqrt{3}}{4}r^2 \Rightarrow \dfrac{r^2}{12}[4 \pi - 3 \sqrt{3}]$
$= \dfrac{R^2}{12} [4 \pi - 3 \sqrt{3}]$