数学知识交流 － 正多边形的面积、周界计算 (2)

Question

如题所述

Answer 1

(1)∵是一正五边形 ∴AB=BC=CD ∴∠ABC=(5-2)x180∘/ 5 =108∘ AB=BC ∠BCA=(180∘- 108∘)/2=36∘ AB/sin∠BCA=AC/sin∠ABC AB=2sin36∘/ sin108∘ 周界=5AB=6.18 (cm) ∠DCA=108∘- ∠BCA =72∘△ADC面积=CD x AC sin∠DCA /2 =1.18 (cm^2) 正五边形面积=1.18+ AB x BC sin108∘/2 X 2 =2.63 (cm^2) (2)∵是一正五边形 ∴AB=BC=CD ∴∠ABC=(5-2)x180∘/ 5 =108∘ AB=BC ∠BCA=(180∘- 108∘)/2=36∘=∠DBC ∠CEB=180∘- ∠BCE - ∠EBC (△内角和) =108∘ BC/sin∠CEB= BE /sin∠BCE BC=sin108∘/ sin36∘=1.62 (cm) 周界=5BC=8.9 (cm) AC/sin∠ABC= AB/sin∠ACBAC=1.62sin108∘/sin 36∘=2.62 (cm) ∠DCA=108∘- ∠BCA =72∘△ADC面积= CD x AC sin∠DCA /2 =2.0143(cm^2)正五边形面积=2.0143+ AB x BC sin108∘/2 X 2 =4.50 (cm^2) 2012-04-11 20:08:54 补充： (2)BC=sin108∘/ sin36∘=1.62 (cm) 周界=5BC=8.09 (cm) 岩先打小个0
参考: me o_o
(1)let a be AB=BC ∠ABC=(5-2)*180/5=108 a^2+a^2-2(a)(a)cos108=AC^2 Cosine theorem: 2a^2-2a^2 cos 108=4 2a^2(1- cos 108)=4 a^2=2/(1-cos 108) a=√[2/(1-cos 108)] ----------------------------------- 正多边形面积公式: (5){√[2/(1-cos 108)]}^2 / 4tan (180/5) =5*[2/(1-cos 108)] / 4tan 36 =2.63 sq.units (corr.to 3sig.fig.) ------------------------------------ 周界=5a =5√[2/(1-cos 108)] =6.18units(corr.to 3sig.fig.) ---------------------------------------- (2)let a be BC ∠CEB=∠ABC (用个 4边形同vert. opp.∠) ∠ABC=(5-2)*180/5=108 cosine theorem: 1^2+1^2-2(1)(1)cos 108=a^2 2-2cos 108=a^2 a=√(2-2cos108) ---------------------------------------- 正多边形面积公式: (5)[√(2-2cos108)]^2 / 4tan(180/5) =5*(2-2cos108) / 4tan 36 =4.50sq.units(corr.to 3sig.fig.) ---------------------------------------- 周界=5a =5√(2-2cos108) =8.09units(corr.to 3sig.fig.)