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Join all vertices to the midpoint of the 15-agon. This will form 15 congruent isosceles triangles with an angle of 360° ÷ 15 = 24° at the centre and a base of length equal to that of the 15-agon = 360cm ÷ 15 = 24 cm

The area of the 15-agon is the sum of the areas of these triangles = 15 × area one triangle

Drop a perpendicular form the centre of the 15-agon to one side and it divides the isosceles triangle into two right angle triangles with an angle half that of the isosceles triangle at the centre of the 15-agon with one leg the height of the triangle and the other leg equal to half the side of the 15-agon

Using Trigonometry, the height of the triangle, and thus its area, and hence the area of the 15-agon, can be found.

height = ½ side / tan ½ angle at centre

→ area 15-agon = 15 × ½ × side × ½ side / tan ½ angle

= 15 × ¼ × side² ÷ tan ½ angle

= 15/4 × (24 cm)² ÷ tan(½×360°÷15)

≈ 10162 cm²

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Using the above it can be seen for a regular n-agon, the area is given by:

area = n × side² ÷ (4 × tan(180° ÷ n))

= perimeter² ÷ (4 × n × tan(180° ÷ n))

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