(1)∵ Is it a regular Pentagon ∴ AB = BC = CD ∴∠ ABC = (5-2) x180? / 5 = 108? AB=BC ∠BCA=( 180？ - 108? )/2=36? AB/sin∠BCA=AC/sin∠ABC AB=2sin36？ / sin 108？ Circumference = 5ab = 6.18 (cm) ≈ DCA =108? - ∠BCA =72？ △ADC area = CD x AC sin ∠ DCA/2 =1.18 (cm2) regular pentagonal area =1.18+ab x bcsin108? /2x2 = 2.63 (cm 2) (2) ∵ Is it a regular Pentagon ∴ AB = BC = CD ∴∠ ABC = (5-2) x180? / 5 = 108? AB=BC ∠BCA=( 180？ - 108? )/2=36? =∠DBC ∠CEB= 180？ -∠ BCE-∠ EBC (sum of delta internal angles) = 108? BC/sin∠CEB = BE/sin∠BCE BC = sin 108？ / sin36？ = 1.62 (cm) perimeter = 5bc = 8.9 (cm) ac/sin ∠ ABC = ab/sin ∠ a cbac =1.62 sin108? /sin 36？ =2.62 (cm) ∠DCA= 108？ - ∠BCA =72？ △ADC area = CD x AC sin ∠ DCA/2 = 2.0143 (cm2) Regular Pentagon area = 2.0143+ab x bcsin108? /2x2 = 4.50 (cm 2) 2012-04-120: 08: 54 Supplement: (2)BC=sin 108? / sin36？ = 1.62 (cm) perimeter =5BC=8.09 (cm) rock, make a small 0 first.

Reference: me o_o

(1) let a be ab = BC ∠ ABC = (5-2) *180/5 =108a 2+a 2-2 (a) (a) cos108. -cos108) = 4a 2 = 2/(1-cos108) a = √ [2/(1-cos108)]-Regular polygon area formula: (. Unit (corrected to 3sig. Figure.)-perimeter = 5a = 5 √ [2 = 6. 18 units (corr.to3sig.fig.)-(2) Let A be BC. 5= 108 cosine theorem:12+12 (1) (1) cos108 = a 22-2 cos/kloc-0. 4 Tan 36 = 4.50 square. Unit (corrected to 3sig. Figure)-perimeter = 5a =