The area of the considered octagon which has distance from its center to the vertex of 9 m is given by: Option b). 229.1 m²
How to find the area of a regular octagon?
If the regular octagon has sides of 'a' units, then its area is:
[tex]A = 2(1+\sqrt{2})a^2 \: \rm unit^2[/tex]
Consider the attached figure below (second figure too).
If we find the value of 'a', then we can use the above formula.
Since there are 8 equal sides in a regular octagon, the angle from center to each of the side is full angle / 8 = 360/8 = 45 degrees.
Since the angle OA is perpendicular on BC, due to symmetry, we have the angle BOC divided into 2 parts.
Thus, we get:
m∠AOB = 45/2 degrees = 22.5°
Using the sine ratio from the perspective of angle AOB, we get:
[tex]\sin(m\angle AOB) = \dfrac{|AB|}{|OB|}\\\\\sin(22.5^\circ) = \dfrac{a/2}{9}\\\\a \approx 2 \times 9 \times 0.38268 \approx 6.888[/tex] meters
(since length of AB is half of length of BC, which is 'a' units).
Thus, we get:
[tex]A = 2(1+\sqrt{2})a^2 \approx 2(1+\sqrt{2})(6.888)^2 \approx 229.1\: \rm m^2[/tex]
Thus, the area of the considered octagon which has distance from its center to the vertex of 9 m is given by: Option b). 229.1 m²
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