найдем радиус вписанной окружности
r=a/(2 tg pi/12)
S=12*r*a
S=12*a^2/(2 tg pi/12)=12 36 (2-sqrt(3))/(2 tg pi/12)=216 (2-sqrt(3))/ ( 1/2 / 1+sqrt(3)/2))=
216 (2-sqrt(3))/(2+sqrt(3))=216 (2-sqrt(3))^2=216 (4+3+4sqrt(3))=216 (7+4sqrt(3))
найдем радиус вписанной окружности
r=a/(2 tg pi/12)
S=12*r*a
S=12*a^2/(2 tg pi/12)=12 36 (2-sqrt(3))/(2 tg pi/12)=216 (2-sqrt(3))/ ( 1/2 / 1+sqrt(3)/2))=
216 (2-sqrt(3))/(2+sqrt(3))=216 (2-sqrt(3))^2=216 (4+3+4sqrt(3))=216 (7+4sqrt(3))