✔ 最佳答案
圖片參考:
https://s.yimg.com/lo/api/res/1.2/sPMz8WVf1WjSBbG.fF_k8g--/YXBwaWQ9dHdhbnN3ZXJzO3E9ODU-/https://sp.yimg.com/ib/th?id=JN.lzQ3Ki7mlvnqVcUC58pbfg&pid=15.1&P=0
圖中△CPB ≌ △AQB , 則 BQ = PB = 2, ∠PBQ是直角, 得
PQ = √(2² + 2²) = √8 , 而 AQ = 3 , PA = 1 , 注意成立
PA² + PQ² = AQ² , 故 ∠APQ為直角。
於是∠ APB =∠APQ +∠BPQ = 90° + 45° = 135°.
在△APB中由餘弦定理得正方形面積
= AB² = 1² + 2² - 2(1)(2)cos135° = 5 - 4(- √2/2) = 5 + 2√2
2015-05-03 06:44:23 補充:
C
P
A____________B
Q
AP = 1 , PB = 2 , PC = 3