《2025年学霸题中题八年级数学上册人教版》

1. 如图所示，D为△ABC中AC边上一点，AD=1，DC=2，AB=4，E是AB上一点，且△ABC的面积等于△DEC面积的2倍，则BE的长为（B）

$A. \frac { 1 } { 2 } $
B. 1
$C. \frac { 2 } { 3 } $
$D. \frac { 4 } { 3 } $

答案： B 解析:设 $ S_{\triangle ADE} = S $。$\because AD = 1$，$DC = 2$，$\therefore DC = 2AD$。$\because \triangle CDE$ 和 $\triangle ADE$ 同高，设高为 $ h_1 $，$\therefore S_{\triangle CDE} = \frac{1}{2}CD \cdot h_1 = \frac{1}{2} \times 2AD \cdot h_1 = 2 \times \frac{1}{2}AD \cdot h_1 = 2S_{\triangle ADE} = 2S $。$\because \triangle ABC$ 的面积等于 $\triangle DEC$ 面积的 2 倍，$\therefore S_{\triangle ABC} = 4S$。$\because S_{\triangle CBE} + S_{\triangle ADE} + S_{\triangle CDE} = S_{\triangle ABC}$，$\therefore S_{\triangle CBE} + S + 2S = 4S$，$\therefore S_{\triangle CBE} = S = \frac{1}{4} \times 4S = \frac{1}{4}S_{\triangle ABC}$。$\because AB = 4$，$\triangle ABC$ 和 $\triangle BCE$ 同高，设高为 $ h_2 $，$\therefore \frac{1}{2}BE \cdot h_2 = \frac{1}{4} \times \frac{1}{2}AB \cdot h_2$，$\therefore BE = \frac{1}{4}AB = \frac{1}{4} \times 4 = 1$。故选 B。

2. (1)如图①，AD，BE是△ABC的两条中线，AD与BE交于点O，若△ABO的面积为3，则△ABC的面积为______9。
(2)如图②，D，E分别在△ABC的边BC，AC上，$$ C D = \frac { 1 } { 3 } B C $$，$$ C E = \frac { 1 } { 3 } A C $$，AD与BE交于点O，若△ABC的面积为12，求△ABO的面积。
(3)如图③，△ABC的面积等于35，AE=ED，BD=3DC，求图中阴影部分的面积。15

答案：
(1) 9 解析: 连接 $ OC $，则 $ S_{\triangle BOD} = S_{\triangle COD}$，$ S_{\triangle AOE} = S_{\triangle COE}$，而 $ S_{\triangle ADC} = S_{\triangle BEC} = \frac{1}{2}S_{\triangle ABC}$，$\therefore S_{\triangle ADC} - S_{四边形DOEC} = S_{\triangle BEC} - S_{四边形DOEC}$，即 $ S_{\triangle AOE} = S_{\triangle BOD}$。$\therefore S_{\triangle EOC} = S_{\triangle COD} = S_{\triangle AOE} = S_{\triangle BOD} = \frac{1}{3}S_{\triangle ADC} = \frac{1}{3} \times \frac{1}{2}S_{\triangle ABC} = \frac{1}{6}S_{\triangle ABC}$，$ S_{\triangle AOC} = S_{\triangle BOC} = \frac{1}{3}S_{\triangle ABC}$，$\therefore S_{\triangle ABC} = 3S_{\triangle ABO} = 9$。
(2) 连接 $ OC $。$\because CD = \frac{1}{3}BC$，$ CE = \frac{1}{3}AC$，$\therefore S_{\triangle ACD} = S_{\triangle BCE} = \frac{1}{3}S_{\triangle ABC} = 4$。又 $\because S_{\triangle ACD} - S_{四边形ODCE} = S_{\triangle BCE} - S_{四边形ODCE}$，$\therefore S_{\triangle AOE} = S_{\triangle BOD}$。又 $\because AE : EC = 2 : 1 = BD : DC$，$\therefore S_{\triangle OEC} = \frac{1}{2}S_{\triangle AOE}$，$ S_{\triangle ODC} = \frac{1}{2}S_{\triangle BOD}$，$\therefore S_{\triangle OEC} = S_{\triangle ODC}$。$\therefore S_{\triangle BCE} = S_{\triangle BOD} + S_{\triangle ODC} + S_{\triangle OEC} = 4S_{\triangle ODC} = 4$，即 $ S_{\triangle ODC} = 1$。$\therefore S_{\triangle AOE} = S_{\triangle BOD} = 2$，$\therefore S_{\triangle ABO} = S_{\triangle ABC} - S_{\triangle ADC} - S_{\triangle BOD} = 12 - 4 - 2 = 6$。
(3) 连接 $ DF $，由题意得 $ S_{\triangle AEF} = S_{\triangle EFD}$，$ S_{\triangle ABE} = S_{\triangle BED}$，$ S_{\triangle BDF} = 3S_{\triangle FDC}$，$ S_{\triangle ABD} = 3S_{\triangle ADC}$。设 $\triangle AEF$ 的面积为 $ x $，$\triangle BDE$ 的面积为 $ y $，则 $ x + x + y + y + \frac{1}{3}(x + y) = 35$，解得 $ x + y = 15$，$\therefore$ 阴影部分的面积为 15。

3. 如图①，我们知道若直线$$ l _ { 1 } // l _ { 2 } $$，则△ABC与△ABD的面积相等；反之，若△ABC与△ABD的面积相等，则也可得到直线$$ l _ { 1 } // l _ { 2 } $$，利用此知识解答以下问题：
如图②，已知AB//CD，AD//CB，P，Q分别是线段BC，CD上的点，$$ C P = \frac { 1 } { 3 } B C $$，$$ C Q = \frac { 1 } { 3 } C D $$，E，F分别是线段AB，AD上的点，$$ A E = \frac { 2 } { 3 } A B $$，$$ A F = \frac { 2 } { 3 } A D $$，连接PQ，EF，若△PCQ的面积是4。
(1)求四边形ABCD的面积；
(2)求证：PQ//EF。

答案：

(1) 如图所示，连接 $ AC $，$ BD $ 交于 $ O $，连接 $ PD $。$\because CQ = \frac{1}{3}CD$，$\triangle CPD$ 和 $\triangle CPQ$ 同高（分别以 $ CD $，$ CQ $ 为底）， $\therefore S_{\triangle PCD} = 3S_{\triangle PCQ} = 12$，同理可得 $ S_{\triangle BCD} = 3S_{\triangle PCD} = 36$。$\because AD // BC$，$\therefore S_{\triangle ABC} = S_{\triangle BCD}$，$\therefore S_{\triangle ABO} = S_{\triangle CDO}$，同理可得 $ S_{\triangle ADO} = S_{\triangle BCO}$，$\therefore S_{\triangle ABD} = S_{\triangle CBD} = 36$，$\therefore S_{四边形ABCD} = S_{\triangle ABD} + S_{\triangle CBD} = 72$。
(2) 如图所示，连接 $ BF $，$ DE $，$ BQ $，由
(1) 得 $ S_{\triangle BDP} = \frac{2}{3}S_{\triangle BCD} = 24$，$ S_{\triangle BDQ} = \frac{2}{3}S_{\triangle BCD} = 24$，$\therefore S_{\triangle BDP} = S_{\triangle BDQ}$，$\therefore PQ // BD$，同理可证 $ EF // BD$，$\therefore PQ // EF$。