: 如圖 兩題
: 跪求神人解惑
: 看起來很簡單，但是到底缺了什麼條件
: -----
: Sent from JPTT on my Gigabyte GSmart Maya M1 v2.
28.
如圖：http://ppt.cc/uHXx
在AD線段外找一點E，使△ADE與△ABO全等
在CD線段外找一點F，使△CDF與△CBO全等
因此 ∠EDA+∠ADC+∠CDF = ∠OBA + 90°+∠CBO = 180°
即 E-D-F 三點共線
連接OE線段、OF線段
∠OEA = ∠OAD + ∠DAE = ∠OAD + ∠BAO = ∠BAD = 90°
因此OE線段 = √(3^2+3^2) = 3√2
同理，∠OCF = 90°，OF線段 = √2
在△OEF中，
OE線段 = 3√2、OF線段 = √2、EF線段 = 4
OF^2 + EF^2 = OE^2
因此∠OFE = 90°
正方形ABCD的面積 = 五邊形AOCFE的面積
= △AOE + △EOF + △FOC
= 3*3/2 + 4*(√2)/2 + 1*1/2
= 5 + 2√2
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