Answer:
The most correct option is;
(B) 958.2 ft.²
Step-by-step explanation:
From the question, the dimension of each square = 3 ft.²
Therefore, the length of the sides of the square = √3 ft.
Based on the above dimensions, the dimension of the small semicircle is found by counting the number of square sides ti subtends as follows;
The dimension of the diameter of the small semicircle = 10·√3
Radius of the small semicircle = Diameter/2 = 10·√3/2 = 5·√3
Area of the small semicircle = (π·r²)/2 = (π×(5·√3)²)/2 = 117.81 ft.²
Similarly;
The dimension of the diameter of the large semicircle = 10·√3 + 2 × 6 × √3
∴ The dimension of the diameter of the large semicircle = 22·√3
Radius of the large semicircle = Diameter/2 = 22·√3/2 = 11·√3
Area of the large semicircle = (π·r²)/2 = (π×(11·√3)²)/2 = 570.2 ft.²
Area of rectangle = 11·√3 × 17·√3 = 561
Area, A of large semicircle cutting into the rectangle is found as follows;
![A_{(segment \, of \, semicircle)} = \frac{1}{4} \times (\theta - sin\theta) \times r^2](https://tex.z-dn.net/?f=A_%7B%28segment%20%5C%2C%20of%20%5C%2C%20semicircle%29%7D%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%5Ctimes%20%28%5Ctheta%20-%20sin%5Ctheta%29%20%5Ctimes%20r%5E2)
Where:
![\theta = 2\times tan^{-1}( \frac{The \, number \, of \, vertical \, squrare \, sides \ cut \, by \ the \ large \, semicircle}{The \, number \, of \, horizontal \, squrare \, sides \ cut \, by \ the \ large \, semicircle} )](https://tex.z-dn.net/?f=%5Ctheta%20%3D%202%5Ctimes%20tan%5E%7B-1%7D%28%20%5Cfrac%7BThe%20%5C%2C%20number%20%5C%2C%20of%20%20%5C%2C%20vertical%20%20%5C%2C%20squrare%20%20%5C%2C%20sides%20%20%5C%20cut%20%20%5C%2C%20%20by%20%20%5C%20%20the%20%20%5C%20%20large%20%20%5C%2C%20%20semicircle%7D%7BThe%20%5C%2C%20number%20%5C%2C%20of%20%20%5C%2C%20horizontal%20%5C%2C%20squrare%20%20%5C%2C%20sides%20%20%5C%20cut%20%20%5C%2C%20%20by%20%20%5C%20%20the%20%20%5C%20%20large%20%20%5C%2C%20%20semicircle%7D%20%29)
![\therefore \theta = 2\times tan^{-1}( \frac{10\cdot \sqrt{3} }{5\cdot \sqrt{3}} ) = 2.214](https://tex.z-dn.net/?f=%5Ctherefore%20%5Ctheta%20%3D%202%5Ctimes%20tan%5E%7B-1%7D%28%20%5Cfrac%7B10%5Ccdot%20%5Csqrt%7B3%7D%20%7D%7B5%5Ccdot%20%5Csqrt%7B3%7D%7D%20%29%20%3D%202.214)
Hence;
![A_{(segment \, of \, semicircle)} = \frac{1}{4} \times (2.214 - sin2.214) \times (11\cdot\sqrt{3} )^2 = 128.3 \, ft^2](https://tex.z-dn.net/?f=A_%7B%28segment%20%5C%2C%20of%20%5C%2C%20semicircle%29%7D%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%5Ctimes%20%282.214%20-%20sin2.214%29%20%5Ctimes%20%2811%5Ccdot%5Csqrt%7B3%7D%20%29%5E2%20%3D%20128.3%20%5C%2C%20ft%5E2)
Therefore; t
The area covered by the pavers = 561 - 128.3 + 570.2 - 117.81 = 885.19 ft²
Therefor, the most correct option is (B) 958.2 ft.².