標題:
發問:
四個部份之面積如圖所示。求ΔABC 的面積。 http://pic.fksite.com/080301/205256by_fksite_com.JPG
最佳解答:
As follows: 圖片參考：http://i238.photobucket.com/albums/ff245/chocolate328154/Maths192.jpg?t=1204385817 Suppose all triangles meet at O. Since ΔAOB and ΔBOE have the same height, their ratio of areas is 70 : 35 = 2 : 1. Then AO : OE = 2 : 1. Since ΔAOC and ΔCOE have the same height, their ratio of areas is 2 : 1. 1 / 2 = y / ( 84 + x ) x = 2y - 84 --- ( 1 ) Similarly, by considering same heights, area of ΔCOD : area of ΔCOB = x : 70, 84 / ( y + 35 ) = x / 70 --- ( 2 ) Put ( 1 ) into ( 2 ), 84 / ( y + 35 ) = ( 2y - 84 ) / 70 y2 - 7y - 4410 = 0 By quadratic formula, y = 70 x = 2 ( 70 ) - 84 = 56 Therefore area of ΔABC: 84 + 56 + 40 + 30 + 35 + 70 = 315
其他解答:
84 + 56 + 40 + 30 + 35 + 70 = 315
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math 難題2發問:
四個部份之面積如圖所示。求ΔABC 的面積。 http://pic.fksite.com/080301/205256by_fksite_com.JPG
最佳解答:
As follows: 圖片參考：http://i238.photobucket.com/albums/ff245/chocolate328154/Maths192.jpg?t=1204385817 Suppose all triangles meet at O. Since ΔAOB and ΔBOE have the same height, their ratio of areas is 70 : 35 = 2 : 1. Then AO : OE = 2 : 1. Since ΔAOC and ΔCOE have the same height, their ratio of areas is 2 : 1. 1 / 2 = y / ( 84 + x ) x = 2y - 84 --- ( 1 ) Similarly, by considering same heights, area of ΔCOD : area of ΔCOB = x : 70, 84 / ( y + 35 ) = x / 70 --- ( 2 ) Put ( 1 ) into ( 2 ), 84 / ( y + 35 ) = ( 2y - 84 ) / 70 y2 - 7y - 4410 = 0 By quadratic formula, y = 70 x = 2 ( 70 ) - 84 = 56 Therefore area of ΔABC: 84 + 56 + 40 + 30 + 35 + 70 = 315
其他解答:
84 + 56 + 40 + 30 + 35 + 70 = 315
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