Question

求解比例数学题

如图ABCD是一个矩形,且长与宽的比为3:2,E在BC上,F在CD上,并且三角形ABE三角形ADF,四边形AECF的面积相等,四边形AECF的面积相等,求三角形AEF与矩形ABCD面积之比
https://gss0.baidu.com/9fo3dSag_xI4khGko9WTAnF6hhy/zhidao/wh%3D600%2C800/sign=1a486dea367adab43d851345bbe49f24/c2cec3fdfc03924534d93fcd8794a4c27c1e2580.jpg

Answer 1

∵S△ABE=S△ADF
∴1/2AB*BE = 1/2AD*DF
∴AD/AB=BE/DF
又：AD/AB=3/2
∴BE/DF=3/2

令AD=3x，AB=2x；BE=3y，DF=2y，则有：
FC=DC-DF=AB-DF=2x-2y
EC=BC-BE=AD-BE=3x-3y

∵S△ABE=SAECF=S△AEC+S△ACF
∴1/2AB*BE=1/2EC*AB+1/2FC*AD
即：
1/2*2x*3y = 1/2*(3x-3y)*2x + 1/2*(2x-2y)*3x
y=2(x-y)
2x=3y
y = 2x/3

S△AEF=SAECF-S△ECF=S△ABE-S△EFC=1/2*AB*BE-1/2EC*FC
= 1/2*2x*3y-1/2(3x-3y)*(2x-2y)
= 1/2*2x*2x-1/2(3x-2x)*(2x-4x/3)
= 5²/3
SABCD=AB*AD=3x*2x=6x²
S△AEF/SABCD = 5²/3 /(6x²) = 5/18

Answer 2

-2+3=2乘以-2+5
1=-4+5
1=1
为什么...
-2乘以2=-4
5-4=1
3-2=1
答 在X等于-2的情况下等式计算完1=1
X+3=2X+5 等式成立.