✔ 最佳答案
圖片參考：
http://desmond.imageshack.us/Himg444/scaled.php?server=444&filename=rectangle.jpg&res=landing
1)
△QAP ~ △QCD (A.A.A.)
∴ PQ : DQ = 6 : 9
i.e. DQ : DP = 9 : (9+6) = 3 : 5
△QAR ~ △QCB (A.A.A.)
∴ AR : BC = AQ : QC = 6 : 9
So AR : AD = 6 : 9 since AD = BC ,
i.e. RD : AD = (9-6) : 9 = 1 : 3
△QDR
= △AQD * AR/AD
= (△APD * DQ/DP) * RD/AD
= 6*16/2 * 3/5 * 1/3
= 9.6 cm²
圖片參考：
http://imgcld.yimg.com/8/n/HA04628698/o/701206050028213873410310.jpg
2)
設 E 為 D 至 AB 垂足 , o 為AC , BD 之交點 。
∠DBE = 180° - 15° - 30° - 90° = 45° = ∠BDE
∠DCA = 45 - 30 = 15° = ∠CAB
∠AoB =∠CoD(對頂角)
∴ △AoB ~ △CoD (A.A.)
設 BC = BD = 1 , 則
Do : oB = (1 - tan30°) : tan30° = (1 - √3/3) : √3/3 = √3 - 1
∴
DC : AB = √3 - 1
√2 : AB = √3 - 1
AB = √2 / (√3 - 1)
AE = AB - BE = √2 / (√3 - 1) - √2/2
tan∠ADE = AE/DE = (√2 / (√3 - 1) - √2/2 ) / √2/2 = 2 / (√3 - 1) - 1 = √3
∴∠ADE = 60°
x =∠ADE + ∠EDB = 60 + 45 = 105°
2012-06-05 18:55:12 補充：
1)
△QDR
= △AQD * AR/AD
should be
△QDR
= △AQD * RD/AD
2012-06-05 19:55:58 補充：
You're welcome~
