How do you solve for the area if a regular polygon with a perimeter of 72ft and an interior angle of 170 degrees?
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Interior angle = 170 deg.
Triangle joining center of polygon O, and two vertices A and B - has angles equal to 170/2, 170/2 and 180 - 170 = 10 deg.
AOB is an isosceles triangle.
angle AOB = 10 deg.
hence there are 360/10 = 36 sides in the polygon.
each side = 72 feet / 36 = 2 feet = AB
Altitude of AOB = AB/(2 * tan 5 deg) = cot 5
area of one triangle AOB : 1/2 * cot 5 * 2 = Cot 5 square feet
total area of polygon = 36 * Cot 5 deg square feet
Triangle joining center of polygon O, and two vertices A and B - has angles equal to 170/2, 170/2 and 180 - 170 = 10 deg.
AOB is an isosceles triangle.
angle AOB = 10 deg.
hence there are 360/10 = 36 sides in the polygon.
each side = 72 feet / 36 = 2 feet = AB
Altitude of AOB = AB/(2 * tan 5 deg) = cot 5
area of one triangle AOB : 1/2 * cot 5 * 2 = Cot 5 square feet
total area of polygon = 36 * Cot 5 deg square feet
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