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The area of a regular octagon can be calculated using the formula ( A = 2 \times (1 + \sqrt{2}) \times s^2 ), where ( s ) is the side length. For a side length of 22, the area would be ( A = 2 \times (1 + \sqrt{2}) \times 22^2 ). This results in an area of approximately 1,210.36 square units.

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What is the perimeter of a regular octagon that has an area of 43560 sq ft?

A regular octagon is formed from 8 isosceles triangles, each having a base length of 'd' (the length of a side of the octagon) and base angles of 67.5°. The height of one triangle is, 1/2d.tan67.5 The area of one triangle is, 1/2(base x height) = 1/2d.1/2d.tan67.5 = 1/4.d2tan67.5 The area of the octagon = 8 x area of one triangle = 2d2tan 67.5 = 4.8284d2. Thus, 4.8284d2 = 43560 : d2 = 9021.5714 : d = 94.982 The perimeter = 8d = 759.86 ft. (2dp). Note : It can be shown that the area, A = d2(2 + 2√2).


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