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第35頁

信息發(fā)布者：

$???解：(1)\frac {AB}{BC}=\frac {3}{4}，\frac {CF}{CD}=\frac {4-\frac {7}{4}}{3}=\frac {3}{4}???$

$???(2)因為\frac {AB}{BC}=\frac {CF}{CD}=\frac {3}{4}???$

$???所以線段AB，BC，CF，CD成比例???$

$???解：(1)DB=12-AD??????$

$所以\frac {AD}{12-AD}=\frac {AE}{EC}=\frac {6}{4}??????$

$所以AD=\frac {36}{5}$

$??????(2)DB=12-\frac {36}{5}=\frac {24}{5}，AC=6+4=10$

$??????所以\frac {DB}{AB}=\frac {\frac {24}{5}}{12}=\frac {2}{5}$

$??????\frac {EC}{AC}=\frac {4}{10}=\frac {2}{5}??????$

$所以\frac {DB}{AB}=\frac {EC}{AC}$

$???解：設S_{△COD}= x ， S_{△AOE}=y，???$

$???則\frac {35+x}{30}=\frac {OC}{OF}=\frac {y+84}{40}???$

$???即4x- 3y= 112???$

$???\frac {x}{y+84}=\frac {OD}{OA}=\frac {35}{30+40}???$

$???即2x-y= 84???$

$???解得x= 70，y= 56???$

$???所以S_{△ABC}= 40+ 30+ 35+ 70+84+ 56= 315???$

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