问题
如图,$ABCD$是梯形,面积为$180$,$BC=2AD$,$AE=2DE$,$BF=2FC$,连接$EF$,$W$是$EF$上一点,$WC$将四边形$CDEF$分割成面积相等的两块图形.
(1)求四边形$CDEF$的面积.
(2)求$EW:WF$.
(3)求三角形$ABW$的面积.
解析
(1)因为$AE=2DE$,$BF=2FC$,所以$S_{ABFE}=2S_{CDEF}$,所以$S_{CDEF}=180\times \dfrac{1}{3}=60$.
(2)如图,连接$CE$$$S_{\triangle FWC}=60\div 2=30$$
$$S_{\triangle CDE}=\dfrac{1}{3}S_{\triangle ADC}=\dfrac{1}{3}\times \dfrac{1}{3}S_{ABCD}=20$$
$$S_{\triangle EWC}=30-20=10$$
所以$$EW:WF=S_{\triangle EWC}:S_{\triangle FWC}=10:30=1:3$$
(3)易得
$$S_{ABFE}=180-60=120$$
$$S_{\triangle AEW}=\dfrac{1}{4}S_{\triangle AEF}=\dfrac{1}{4}\times \dfrac{1}{3}\times S_{ABFE}=\dfrac{1}{12}\times 120 =10$$
$$S_{\triangle BFW}=\dfrac{3}{4}S_{\triangle BEF}=\dfrac{3}{4}\times \dfrac{2}{3}\times S_{ABFE}=\dfrac{1}{2}\times 120 =60$$
所以$S_{\triangle ABW}=120-10-60=50$.
暂无评论内容