Question

如图,在四边形ABCD中,∠B+∠D=180°,AB=AD,AC=二倍根号三,∠ACD=60°,求四边形ABCD的面积.

Answer 1

AB/sin∠ACB＝AC/sin∠B,AD/sin∠ACD＝AC/sin∠D

∵∠B﹢∠D＝180°∴sin∠B＝sin∠D

又∵AB＝AD∴∠ACB＝∠ACD＝60°

S＝1/2·BC·AC·sin∠BCA＋1/2·DC·AC·sin∠DCA

∵∠ACB＝∠ACD＝60°,AC＝2倍根号3

∴S＝3/2﹙BC＋DC﹚

cos∠ACD＝﹙AC²＋CD²－AD²﹚／2·AC·CD＝1/2

cos∠ACB＝﹙AC²＋BC²－AB²﹚／2·AC·BC＝1/2

12＋CD²－AD²＝2倍根号3×CD

12＋BC²－AD²＝2倍根号3×BC 两式作差

CD²－BC²＝2倍根号3×﹙CD－BC﹚＝﹙CD＋BC﹚×﹙CD－BC﹚

∴CD＋BC＝2倍根号3

∴S＝3/2·2倍根号3＝3倍根号3