hitunglah luas dari segi-dua belas beraturan, jika panjang sisinya 6 cm.

L segi-n = n/2 . r kuadrat . Sin 360/n
= 12/2 . 6 kuadrat . Sin 360/12
= 6. 36. Sin 30
= 216 . 1/2
= 108

Biar jari jari lingkaran = r Maka, [tex] 2r^2 - 2r^2cos30 = 36 [/tex] Sehingga [tex] r^2(2-\sqrt{3}) = 36 [/tex] Jadi [tex] r^2 = \frac{36}{2-\sqrt{3}} [/tex] Luas =
[tex] \frac{n}{2}. r^2. sin(\frac{360}{n}) [/tex]

=[tex] \frac{12}{2}. \frac{36}{2-\sqrt{3}}. sin(\frac{360}{12}) [/tex]

=[tex] 6. \frac{36}{2-\sqrt{3}} . sin(30) [/tex]
=[tex] \frac{216}{2-\sqrt{3}}. \frac{1}{2} [/tex]
=[tex] ( \frac{108}{2-\sqrt{3}})cm^2 [/tex]