Regular octagon has 8 sides, all of same length. The approximate area of the octagonal window specified, along with the frame is: Option D: 11.2 ft²
We can cut that polygon in pieces such that slices are from center to vertices. Then we will get isosceles triangles (n such triangles, n being the number of sides in that polygon).
Area of polygon = n times area of such isosceles triangles (all are congruent, so same area).
An isosceles triangle has two of its side length of same measure.
Let the considered isosceles triangle has two same lengthed sides of 'a' units.
Let its third side be of 'b' units.
Then, if we drop a perpendicular from the vertex joining similar sides to the opposite side, the opposite side is cut in half, each of length b/2 units.
Using the Pythagoras theorem, we get the length of that perpendicular as:
[tex]L = \sqrt{a^2 - (\dfrac{b}{2})^2} = \dfrac{\sqrt{4a^2 - b^2}}{2} \: \rm units.[/tex]
Thus, area of that triangle = 1/2 times base (b) times height(L)
Thus, [tex]A = \dfrac{1}{2} \times b \times \dfrac{\sqrt{4a^2 - b^2}}{2} = \dfrac{b\sqrt{4a^2 - b^2}}{4} \: \rm unit^2[/tex]
Refer to the figure attached below.
The area of the window = [tex]8 \times A[/tex]
where A is the area of each of the triangle pieces.
The length of the similar side = a = 2 ft.
The length of the third side = b = 1.52 ft.
Thus, the area A of that isosceles triangle is found to be
[tex]A = \dfrac{b\sqrt{4a^2 - b^2}}{4} = \dfrac{1.52 \times \sqrt{4.(2^2) - (1.52)^2}}{4}\\\\A = \dfrac{1.52 \times \sqrt{13.6896}}{4} \approx 1.4059 \: \rm ft^2[/tex]
Thus, area of window = 8 times A ≈ 11.247 ≈ 11.2 ft²
Thus, The approximate area of the octagonal window specified, along with the frame is: Option D: 11.2 ft²
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