Join all vertices to the midpoint of the 15-agon. This will form15 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 cmThe area of the 15-agon is the sum of the areas of thesetriangles = 15 × area one triangleDrop a perpendicular form the centre of the 15-agon to one sideand it divides the isosceles triangle into two right angletriangles with an angle half that of the isosceles triangle at thecentre of the 15-agon with one leg the height of the triangle andthe other leg equal to half the side of the 15-agonUsing Trigonometry, the height of the triangle, and thus itsarea, 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------------------------------------------------------Using the above it can be seen for a regular n-agon, the area isgiven by:area = n × side² ÷ (4 × tan(180° ÷ n))= perimeter² ÷ (4 × n × tan(180° ÷ n))