What is the area of the shaded portion in circle O,

[BTGmoderatorLU]

Source: Veritas Prep



What is the area of the shaded portion in circle O, as pictured above, with ∠O = 60° and radius r = 10?
$$\text{A. } \frac{50}{3}\pi - 50\sqrt{3}$$
$$\text{B. } \frac{50}{3}\pi - 50$$
$$\text{C. } \pi\sqrt{3}$$
$$\text{D. } \frac{50}{3}\pi - 25\sqrt{3}$$
$$\text{E. } \frac{50}{3}\pi$$
The OA is D

[fskilnik@GMATH]

Source: Veritas Prep



What is the area of the shaded portion in circle O, as pictured above, with ∠O = 60° and radius r = 10?
$$\text{A. } \frac{50}{3}\pi - 50\sqrt{3}\,\,\,\,\,\,\text{B. } \frac{50}{3}\pi - 50\,\,\,\,\,\,\text{C. } \pi\sqrt{3}\,\,\,\,\,\,\text{D. } \frac{50}{3}\pi - 25\sqrt{3}\,\,\,\,\,\,\text{E. } \frac{50}{3}\pi$$

$$? = {S_{{\rm{circ}}{\rm{.sector}}}} - {S_{\Delta {\rm{equil}}}}$$
$${S_{{\rm{circ}}{\rm{.sector}}}} = {{60} \over {360}}\left( {\pi \cdot {{10}^2}} \right) = {1 \over 6}\left( {100\pi } \right)$$
$${S_{\Delta {\rm{equil}}}} = {{10 \cdot {h_{{\rm{equil}}}}} \over 2}\mathop = \limits^{\left( * \right)} 5 \cdot 5\sqrt 3 $$
$$\left( * \right)\,\,\,{h_{{\rm{equil}}}}\,\,\,\mathop = \limits^{{\rm{shortcut}}} \,\,\,{{L\sqrt 3 } \over 2}\mathop = \limits^{} 5\sqrt 3 $$
$$? = {{50\pi } \over 3} - 25\sqrt 3 $$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.