標題:
三角形面積
http://i31.tinypic.com/mt4ghf.jpg 如圖所示，ＡＥ＝ＥＤ，ＢＣ＝３ＢＤ， 三角形ＡＢＣ的面積＝３０平方厘米 陰影部份的面積是多少 ＊要用比的方法計算 更新: if DC=1/3BD instead of ＢＣ＝３ＢＤ 更新 2: if DC=1/3BD instead of ＢＣ＝３ＢＤ only required to explain this part
最佳解答:
BC = 3BD BD : DC = 1 : 2 so, Area of ΔABD : area of ΔADC = 1 : 2 (same height) i.e. Area of ΔABD = 10 cm^2 and area of ΔADC = 10 cm^2 AE = ED so, area of ΔEDC = area of ΔAEC (same height) = 10 cm^2 Produce CE to meet AB at F. Then join FD and let area of ΔEFD be x cm^2 Then area of ΔAFE is also x cm^2. Hence, area of ΔFBD is (10 - 2x) cm^2 Area of ΔFBD : area of ΔFDC = BD : DC (same height) = 1 : 2 i.e. (10 - 2x) / (10 + x) = 1/2 which gives x = 2 Therefore, area of the shaded region = (2 + 10) cm^2 = 12 cm^2
其他解答:
三角形面積
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發問:http://i31.tinypic.com/mt4ghf.jpg 如圖所示，ＡＥ＝ＥＤ，ＢＣ＝３ＢＤ， 三角形ＡＢＣ的面積＝３０平方厘米 陰影部份的面積是多少 ＊要用比的方法計算 更新: if DC=1/3BD instead of ＢＣ＝３ＢＤ 更新 2: if DC=1/3BD instead of ＢＣ＝３ＢＤ only required to explain this part
最佳解答:
BC = 3BD BD : DC = 1 : 2 so, Area of ΔABD : area of ΔADC = 1 : 2 (same height) i.e. Area of ΔABD = 10 cm^2 and area of ΔADC = 10 cm^2 AE = ED so, area of ΔEDC = area of ΔAEC (same height) = 10 cm^2 Produce CE to meet AB at F. Then join FD and let area of ΔEFD be x cm^2 Then area of ΔAFE is also x cm^2. Hence, area of ΔFBD is (10 - 2x) cm^2 Area of ΔFBD : area of ΔFDC = BD : DC (same height) = 1 : 2 i.e. (10 - 2x) / (10 + x) = 1/2 which gives x = 2 Therefore, area of the shaded region = (2 + 10) cm^2 = 12 cm^2
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