pho chopstix

j Pho
Pho ()7.8.
pho 7 PhU+2C5BE 2 1 3
Pho Pho .
pho, Pho real, Pho Asia
2011 1 .
pho?
: [f] photography [f'tgrf] 313 5535.
Pho Pho .

j Pho
Pho ()7.8.
pho 7 PhU+2C5BE 2 1 3
Pho Pho .
pho, Pho real, Pho Asia
2011 1 .
pho?
: [f] photography [f'tgrf] 313 5535.
Pho Pho .
Magnitudes:
|ec{OA}| = \sqrt{(-1 + \sqrt{7})^2 + (1 + \sqrt{7})^2} = \sqrt{(1 - 2\sqrt{7} + 7) + (1 + 2\sqrt{7} + 7)} = \sqrt{16 + 8} = \sqrt{24} = 2\sqrt{6}
|ec{OB}| = \sqrt{(-1 - \sqrt{7})^2 + (1 - \sqrt{7})^2} = \sqrt{(1 + 2\sqrt{7} + 7) + (1 - 2\sqrt{7} + 7)} = \sqrt{16 + 8} = \sqrt{24} = 2\sqrt{6}
\cos heta = rac{-12}{(2\sqrt{6})(2\sqrt{6})} = rac{-12}{24} = -rac{1}{2} \Rightarrow heta = rac{2\pi}{3}
The area of the circular sector is:
A_{ ext{sector}} = rac{1}{2} r^2 heta = rac{1}{2} \cdot 16 \cdot rac{2\pi}{3} = rac{16\pi}{3}
Area of triangle \(OAB\):
A_{ ext{triangle}} = rac{1}{2} ab \sin heta = rac{1}{2} (2\sqrt{6})(2\sqrt{6}) \sin\left(rac{2\pi}{3}
ight) = rac{1}{2} \cdot 24 \cdot rac{\sqrt{3}}{2} = 6\sqrt{3}
Area of segment (the smaller region bounded by the chord and arc) is: