What is the area of a sector when r=5/6 and θ=2π/3 radians?

Answer: [tex]\frac{25\pi}{108}[/tex]
Work Shown
[tex]A = \text{area of sector}\\\\A = \frac{\text{angle in radians}}{2\pi}*(\text{area of full circle})\\\\A = \frac{\text{angle in radians}}{2\pi}*\pi*r^2\\\\A = \frac{\text{angle in radians}}{2}*r^2\\\\A = \frac{1}{2}*(\text{angle in radians})*r^2\\\\A = \frac{1}{2}*\frac{2\pi}{3}*\left(\frac{5}{6}\right)^2\\\\A = \frac{1}{2}*\frac{2\pi}{3}*\frac{25}{36}\\\\A = \frac{1*2\pi*25}{2*3*36}\\\\A = \frac{50\pi}{216}\\\\A = \frac{25\pi}{108}[/tex]