Answer:
The area of the sector BOC is A' = 31.4 [tex]cm^{2}[/tex]
Step-by-step explanation:
Radius of the circle = 10 cm = 0.1 m
Angle of sector [tex]\theta[/tex] = 36°
Area of the circle A = [tex]\pi r^{2}[/tex]
A = 3.14 × [tex]0.1^{2}[/tex] = 0.0314 [tex]m^{2}[/tex]
Now area of the sector BOC is
[tex]A' = A (\frac{\theta}{360} )[/tex]
[tex]A' = 0.0314 (\frac{36}{360} )[/tex]
A' = 0.00314 [tex]m^{2}[/tex] = 31.4 [tex]cm^{2}[/tex]
Therefore the area of the sector BOC is A' = 31.4 [tex]cm^{2}[/tex]
