第1个回答 2012-07-21
将ABC置于坐标系中,C为原点,B在+x轴上。 设B(b, 0), A(p, a)
三角形ABC的面积 = (1/2)ab = 240
ab = 480
D((p+b)/2, a/2), E(2b/3, 0)
CD的斜率 = (a/2)/[(p+b)/2] = a/(p+b)
CD的方程: y = ax/(p+b) (1)
AE的方程: (y -0)/(x -2b/3) = (a -0)/(p - 2b/3)
y = a(3x - 2b)/(3p-2b) (2)
联立(1)(2), 可得交点F(2(p+b)/5, 2a/5)
三角形CEF的面积S1 = (1/2)*CE*F的纵坐标
= (1/2)(2b/3)(2a/5)
= 2ab/15 = 2*480/15
= 64
AB的方程: (y - 0)/(x - b) = (a - 0)/(p - b)
ax - (p-b)y - ab = 0
F与AB的距离h = |a*2(p+b)/5 -(p-b)*2a/5 - ab|/√[a² + (p-b)²]
= ab/{5√[a² + (p-b)²]}
AD = √{[p - (p+b)/2]² + (a - a/2)²}
= (1/2)√[a² + (p-b)²]
三角形ADF的面积S2 = (1/2)*AD*F与AB的距离
= (1/2)(1/2)√[a² + (p-b)²]*ab/{5√[a² + (p-b)²]}
= ab/20
= 480/20
= 24
阴影部分的面积=S1 + S2 = 64+24 = 88