《2025年通城学典课时作业本九年级数学下册苏科版江苏专版》

8. （2024·乐山改编）如图，在梯形ABCD中，AD//BC，对角线AC和BD交于点O. 若$\frac{S_{\triangle ABD}}{S_{\triangle BCD}}=\frac{1}{3}$，则$\frac{S_{\triangle AOD}}{S_{\triangle BOC}}$的值为__________，$\frac{C_{\triangle AOD}}{C_{\triangle BOC}}$的值为__________.

答案： $\frac{1}{9}$ $\frac{1}{3}$

9. 如图，DE是△ABC的中位线，F为DE的中点，连接AF并延长，交BC于点G. 若$S_{\triangle EFG}=1$，则$S_{\triangle ABC}$的值为__________.

答案： 24 解析: $\because DE$ 是 $\triangle ABC$ 的中位线, $\therefore DE // AC$. 取 $AG$ 的中点 $M$, 连接 $DM$, 证 $\triangle DFM \cong \triangle EFG$, 得 $MF = GF = \frac{1}{2}GM = \frac{1}{2}AM, S_{\triangle DFM} = S_{\triangle EFG} = 1. \therefore S_{\triangle ADM} = 2S_{\triangle DFM} = 2$. 证 $\triangle ADM \backsim \triangle ABG$, 得 $S_{\triangle ABG} = 4S_{\triangle ADM} = 8$. 证 $\triangle EFG \backsim \triangle CAG$, 得 $S_{\triangle CAG} = 16S_{\triangle EFG} = 16. \therefore S_{\triangle ABC} = S_{\triangle ABG} + S_{\triangle CAG} = 24$.

10. 如图，△ABC的周长为1，连接△ABC三边的中点，得到△A₁B₁C₁，再连接△A₁B₁C₁三边的中点，得到△A₂B₂C₂，…. 若△A₁B₁C₁的周长为$\frac{1}{2}$，则△A₂₀₂₄B₂₀₂₄C₂₀₂₄的周长为__________.

答案： $\frac{1}{2^{2024}}$

11. 如图，E、F是□ABCD对角线AC上的两点，AE = CF = $\frac{1}{4}$AC. 连接DE、DF并延长，分别交AB、BC于点G、H，连接GH. 求$\frac{S_{\triangle ADG}}{S_{\triangle BGH}}$的值.

答案： 连接 $BD$. $\because AE = \frac{1}{4}AC, \therefore \frac{AE}{CE} = \frac{1}{3}$. $\because$ 四边形 $ABCD$ 是平行四边形, $\therefore BA // CD, BA = CD, S_{\triangle ABD} = S_{\triangle BAC} = \frac{1}{2} \cdot S_{\square ABCD}$. $\therefore$ 易得 $\triangle AEG \backsim \triangle CED. \therefore \frac{AE}{CE} = \frac{AG}{CD} = \frac{1}{3}. \therefore \frac{AG}{BA} = \frac{1}{3}. \therefore \frac{BG}{BA} = \frac{2}{3}, S_{\triangle ADG} = \frac{1}{3}S_{\triangle ABD}$. 同理, 可得 $\frac{BH}{BC} = \frac{2}{3}$.
$\therefore \frac{BG}{BA} = \frac{BH}{BC}$. $\because \angle GBH = \angle ABC, \therefore \triangle BGH \backsim \triangle BAC$.
$\therefore \frac{S_{\triangle BGH}}{S_{\triangle BAC}} = (\frac{BG}{BA})^2 = \frac{4}{9}. \therefore S_{\triangle BGH} = \frac{4}{9}S_{\triangle BAC} = \frac{4}{9}S_{\triangle ABD}$.
$\therefore \frac{S_{\triangle ADG}}{S_{\triangle BGH}} = \frac{\frac{1}{3}S_{\triangle ABD}}{\frac{4}{9}S_{\triangle ABD}} = \frac{1}{3} \times \frac{9}{4} = \frac{3}{4}$

12. 如图，△ABC和△DEC的面积相等，点E在边BC上，DE//BA交AC于点F，BA = 12，EF = 9. 求DF的长.

答案： $\because DE // BA, \therefore \triangle FEC \backsim \triangle ABC. \therefore \frac{S_{\triangle FEC}}{S_{\triangle ABC}} = (\frac{EF}{BA})^2 = (\frac{9}{12})^2 = \frac{9}{16}$. $\because \triangle ABC$ 和 $\triangle DEC$ 的面积相等, $\therefore \frac{S_{\triangle FEC}}{S_{\triangle DEC}} = \frac{9}{16}$.
又 $\because \triangle FEC$ 的边 $EF$ 上的高与 $\triangle DEC$ 的边 $DE$ 上的高相同,
$\therefore$ 结合三角形的面积公式, 得 $\frac{EF}{DE} = \frac{9}{16}. \because EF = 9, \therefore DE = 16$.
$\therefore DF = DE - EF = 16 - 9 = 7$