find the area of a circle cirbumscribed about a regular octagon with a perimeter of 80 inches.
+33616 Length of one side of the octagon must be 80/8 = 10 inches.
Angle of one side subtended at the centre = 360°/8 = 45°
Length of side of isosceles triangle formed by lines from centre to one side of the octagon = 5/sin(45/2°) inches. This is also the radius of the circumscribed circle, so:
Area of circle = pi*(5/sin(22.5°))2 in2
$${\mathtt{Area}} = {\mathtt{\pi}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{5}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{22.5}}^\circ\right)}}}\right)}^{{\mathtt{2}}} \Rightarrow {\mathtt{Area}} = {\mathtt{536.303\: \!412\: \!267\: \!149\: \!253\: \!5}}$$
Area ≈ 536.3 in2
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+33616 Length of one side of the octagon must be 80/8 = 10 inches.
Angle of one side subtended at the centre = 360°/8 = 45°
Length of side of isosceles triangle formed by lines from centre to one side of the octagon = 5/sin(45/2°) inches. This is also the radius of the circumscribed circle, so:
Area of circle = pi*(5/sin(22.5°))2 in2
$${\mathtt{Area}} = {\mathtt{\pi}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{5}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{22.5}}^\circ\right)}}}\right)}^{{\mathtt{2}}} \Rightarrow {\mathtt{Area}} = {\mathtt{536.303\: \!412\: \!267\: \!149\: \!253\: \!5}}$$
Area ≈ 536.3 in2
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