Question

三角形ABC的面积是200平方厘米,E在AC上,点D在BC上,且AE:EC=3:5,BD:DC=2:3,AD与BE交于点

Answer 1

∵BD:DC=2:3,SΔABC=200,

∴SΔABD=80,SΔACD=120

∵BD:DC=2:3,AE:EC=3:5,

如图,SΔABF=80-2a

得(80-2a+3b)/(2a+3a+5b)=3/5

80-2a+3b= 3a+3b

a=16

又∵3a+8b=120

48+8b=120

b=9

S四边形DFEC=3a+5b=3*16+5*9=93

所以四边形DFEC的面积为93.