Given:
8 tiles
radius from the center of the archway to the inner edge of the tile. 7 ft.
radius from the center of the archway to the outer edge of the tile. 8 ft = 7 ft + 1 ft.
Area of a semi circle = π r² / 2
A = (3.14 * (7ft)²) / 2 = (3.14 * 49ft²) / 2 = 153.86 ft² / 2 = 76.93 ft²
A = (3.14 * (8ft)²) / 2 = (3.14 * 64ft²) / 2 = 200.96 ft² / 2 = 100.48 ft²
100.48 ft² - 76.93 ft² = 23.55 ft²
23.55 ft² / 8 tiles = 2.94 ft² per tile.
Answer:
It is the first and the last three.
Answer:

Explanation:
The figure is not a regular hexagon. It is an irregular hexagon.
Please, find attached the picture with the original question and the figure.
You can split the figure into two triangles and one rectangle.
The rectangle has dimensions: 7units × 4units, thus its area is 28 units².
Both the upper triangle and lower triangle have base 7 units and height 2 units.
Hence the area of each triangle is:

Hence, the area of the hexagon is:

So, I came up with something like this. I didn't find the final equation algebraically, but simply "figured it out". And I'm not sure how much "correct" this solution is, but it seems to work.
![f(x)=\sin(\omega(x))\\\\f(\pi^n)=\sin(\omega(\pi^n))=0, n\in\mathbb{N}\\\\\\\sin x=0 \implies x=k\pi,k\in\mathbb{Z}\\\Downarrow\\\omega(\pi^n)=k\pi\\\\\boxed{\omega(x)=k\sqrt[\log_{\pi} x]{x},k\in\mathbb{Z}}](https://tex.z-dn.net/?f=f%28x%29%3D%5Csin%28%5Comega%28x%29%29%5C%5C%5C%5Cf%28%5Cpi%5En%29%3D%5Csin%28%5Comega%28%5Cpi%5En%29%29%3D0%2C%20n%5Cin%5Cmathbb%7BN%7D%5C%5C%5C%5C%5C%5C%5Csin%20x%3D0%20%5Cimplies%20x%3Dk%5Cpi%2Ck%5Cin%5Cmathbb%7BZ%7D%5C%5C%5CDownarrow%5C%5C%5Comega%28%5Cpi%5En%29%3Dk%5Cpi%5C%5C%5C%5C%5Cboxed%7B%5Comega%28x%29%3Dk%5Csqrt%5B%5Clog_%7B%5Cpi%7D%20x%5D%7Bx%7D%2Ck%5Cin%5Cmathbb%7BZ%7D%7D)
75 - 75 = 0, therefore, your answer is equal to zero
hope this helps