9. 易错题 [2026成都武侯区段考] 两个相似三角形面积比是4:9，其中一个三角形的周长为24 cm，则另一个三角形的周长是（）

A.16 cm
B.16 cm或28 cm
C.36 cm
D.16 cm或36 cm

9 D $\because$ 两个相似三角形面积比是4:9,$\therefore$ 两个相似三角形相似比是2:3,$\therefore$ 两个相似三角形周长比是2:3. 设另一个三角形的周长是x cm. 两个三角形相似比的顺序未确定,需要分类讨论:①当24: x = 2: 3时,解得x = 36;②当x: 24 = 2: 3时,解得x = 16. 综上,另一个三角形的周长是16 cm或36 cm.

10. [2026宣城期末] 将一副三角板按图叠放，则△AOD与△BOC的面积之比为（）

A.$1:\sqrt{3}$
B.$1:3$
C.$1:\sqrt{2}$
D.$1:2$

10 B 由题意,得$∠ DAB = ∠ ABC = 90°$,$∠ D = 45°$,$∠ C = 30°$,$\therefore ∠ ABD = ∠ D = 45°$,$\therefore AB = AD$,$AC = 2AB$,$\therefore CB = \sqrt{AC^2 - AB^2} = \sqrt{(2AB)^2 - AB^2} = \sqrt{3} AB = \sqrt{3} AD$,$\therefore \frac{AD}{CB} = \frac{AD}{\sqrt{3} AD} = \frac{\sqrt{3}}{3}$. $\because ∠ DAB + ∠ ABC = 180°$,$\therefore AD//CB$,$\therefore △ AOD ∽ △ BOC$,$\therefore \frac{S_{△ AOD}}{S_{△ BOC}} = (\frac{AD}{CB})^2 = (\frac{\sqrt{3}}{3})^2 = \frac{1}{3}$,$\therefore △ AOD$与$△ BOC$的面积之比为1: 3.

11. [2024 皖东南四校联考]如图,将$△ ABC$沿$BC$边上的中线$AD$平移到$△ A'B'C'$的位置.已知$△ ABC$的面积为16,阴影部分三角形的面积为9.若$AA'=1$,则$A'D$等于 ()

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

11 B 如图,设$A'B'$交$BC$于点E,$A'C'$交$BC$于点F. 中线的性质(等分面积)→ $S_{△ A'DE} = \frac{1}{2} S_{△ A'EF} = \frac{9}{2}$, $S_{△ ABD} = \frac{1}{2} S_{△ ABC} = 8$. 平移→$A'E//AB$→$△ DA'E ∽ △ DAB$→$(\frac{A'D}{AD})^2 = \frac{S_{△ A'DE}}{S_{△ ABD}}$→$(\frac{A'D}{A'D + 1})^2 = \frac{\frac{9}{2}}{8} = \frac{9}{16}$→$A'D = 3$ 或$A'D = -\frac{3}{7}$(舍去).

12. [2025合肥瑶海区期中]观察与发现:如图,小明将一个边长为6 cm的正方形纸片ABCD折叠,使得点D落在AB边上的点E处(不与点A,B重合),折痕FH交AD于点F,交BC于点H,点C落在点Q处,EQ与BC交于点G.
(1)小明认为$△ AEF ∽ △ BGE$,你同意吗?请说明理由.
(2)当$AE=2\ \mathrm{cm}$时,请你计算$△ BGE$的周长.

12 解:(1)同意.理由如下:
由折叠的性质,可得$∠ FEQ = ∠ D = 90°$.
$\therefore ∠ AEF + ∠ BEG = 90°$.
$\because ∠ A = 90°$,$\therefore ∠ AEF + ∠ AFE = 90°$,
$\therefore ∠ AFE = ∠ BEG$.
又$\because ∠ A = ∠ B = 90°$,$\therefore △ AEF ∽ △ BGE$.
(2)设$AF = x$ cm,则$EF = DF = (6 - x)$ cm.
在Rt$△ AEF$中,由勾股定理,得$AE^2 + AF^2 = EF^2$,
即$2^2 + x^2 = (6 - x)^2$,$\therefore x = \frac{8}{3}$,即$AF = \frac{8}{3}$ cm.
由折叠的性质,可得$EF = DF$,
$\therefore C_{△ AEF} = AE + AF + EF = AE + AF + DF = AE + AD = 2 + 6 = 8$(cm).
$\because △ AEF ∽ △ BGE$,$\therefore C_{△ AEF}: C_{△ BGE} = AF: BE = \frac{8}{3}: (6 - 2) = 2: 3$,
$\therefore C_{△ BGE} = 8 × \frac{3}{2} = 12$(cm).

13. 运算能力 [2025广安中考] 已知$△ ABC$的面积是1.
(1)如图1,若$D,E$分别是边$BC$和$AC$的中点,$AD$与$BE$相交于点$F$,则四边形$CDFE$的面积为.
(2)如图2,若$M,N$分别是边$BC$和$AC$上距离$C$点最近的六等分点,$AM$与$BN$相交于点$G$,则四边形$CMGN$的面积为.

13 (1)$\boldsymbol{\frac{1}{3}}$;(2)$\boldsymbol{\frac{1}{21}}$
解题思路:(1)连接DE,可证明DE是△ABC的中位线,得到$DE//AB$,$DE = \frac{1}{2} AB$,证明$△ CDE ∽ △ CBA$,可得$\frac{S_{△ CDE}}{S_{△ CBA}} = (\frac{DE}{AB})^2 = \frac{1}{4}$,则$S_{△ CDE} = \frac{1}{4}$,进而可得$S_{△ BDE} = S_{△ CDE} = \frac{1}{4}$,证明$△ DEF ∽ △ ABF$,得到$\frac{EF}{BF} = \frac{DE}{AB} = \frac{1}{2}$,则可得到$\frac{S_{△ DEF}}{S_{△ BDE}} = \frac{EF}{BE} = \frac{1}{3}$,则$S_{△ DEF} = \frac{1}{12}$,据此即可得解;
(2)连接MN,证明$△ CMN ∽ △ CBA$,得到$\frac{S_{△ CMN}}{S_{△ ABC}} = (\frac{CM}{BC})^2 = \frac{1}{36}$,$\frac{MN}{AB} = \frac{CM}{BC} = \frac{1}{6}$,$∠ CMN = ∠ CBA$,则可证明$MN//AB$,$S_{△ CMN} = \frac{1}{36}$,再证明$\frac{S_{△ BMN}}{S_{△ CMN}} = \frac{BM}{CM} = 5$,得到$S_{△ BMN} = \frac{5}{36}$,证明$△ MNG ∽ △ ABG$,得到$\frac{NG}{BG} = \frac{MN}{AB} = \frac{1}{6}$,则$\frac{S_{△ MNG}}{S_{△ BMN}} = \frac{NG}{BN} = \frac{1}{7}$,则$S_{△ MNG} = \frac{5}{252}$,据此即可得解.
(1),连接DE. $\because D,E$分别是边BC和AC的中点,$\therefore DE$是$△ ABC$的中位线,$\therefore DE//AB$,$DE = \frac{1}{2} AB$,$\therefore △ CDE ∽ △ CBA$,$\therefore \frac{S_{△ CDE}}{S_{△ CBA}} = (\frac{DE}{AB})^2 = (\frac{\frac{1}{2} AB}{AB})^2 = \frac{1}{4}$. $\because △ ABC$的面积是1,$\therefore S_{△ CDE} = \frac{1}{4}$. $\because D$是BC的中点,$\therefore S_{△ BDE} = S_{△ CDE} = \frac{1}{4}$. $\because DE//AB$,$\therefore △ DEF ∽ △ ABF$,$\therefore \frac{EF}{BF} = \frac{DE}{AB} = \frac{\frac{1}{2} AB}{AB} = \frac{1}{2}$, $\therefore BF = 2EF$,$\therefore BE = BF + EF = 3EF$,$\therefore \frac{S_{△ DEF}}{S_{△ BDE}} = \frac{EF}{BE} = \frac{EF}{3EF} = \frac{1}{3}$,$\therefore S_{△ DEF} = \frac{1}{12}$,$\therefore S_{四边形CDFE} = S_{△ DEF} + S_{△ CDE} = \frac{1}{12} + \frac{1}{4} = \frac{1}{3}$.
(2),连接MN. $\because M,N$分别是边BC和AC上距离C点最近的六等分点,$\therefore CM = \frac{1}{6} BC$,$CN = \frac{1}{6} AC$,$\therefore \frac{CM}{CB} = \frac{CN}{CA} = \frac{1}{6}$. 又$\because ∠ C = ∠ C$,$\therefore △ CMN ∽ △ CBA$,$\therefore \frac{S_{△ CMN}}{S_{△ ABC}} = (\frac{CM}{BC})^2 = (\frac{\frac{1}{6} BC}{BC})^2 = \frac{1}{36}$,$\frac{MN}{AB} = \frac{CM}{BC} = \frac{\frac{1}{6} BC}{BC} = \frac{1}{6}$, $∠ CMN = ∠ CBA$, $\therefore MN//AB$. $\because △ ABC$的面积是1,$\therefore S_{△ CMN} = \frac{1}{36}$. $\because M$是BC靠近点C的六等分点,$\therefore \frac{BM}{CM} = 5$,$\therefore \frac{S_{△ BMN}}{S_{△ CMN}} = \frac{BM}{CM} = 5$,$\therefore S_{△ BMN} = \frac{5}{36}$. $\because MN//AB$,$\therefore △ MNG ∽ △ ABG$,$\therefore \frac{NG}{BG} = \frac{MN}{AB} = \frac{1}{6}$,$\therefore BG = 6NG$,$\therefore BN = BG + NG = 7NG$,$\therefore \frac{S_{△ MNG}}{S_{△ BMN}} = \frac{NG}{BN} = \frac{NG}{7NG} = \frac{1}{7}$, $\therefore S_{△ MNG} = \frac{5}{252}$,$\therefore S_{四边形CMGN} = S_{△ MNG} + S_{△ CMN} = \frac{5}{252} + \frac{1}{36} = \frac{1}{21}$.

三角形纸片ABC，点D，E分别在边AB，AC上，AD=2DB，沿DE将△ABC剪成面积相等的两部分，则$\frac{AE}{EC}$的值为（）

A.1
B.2
C.3
D.4

1 C 如图,过点D作$DF//BC$交AC于点F(作平行线构造相似三角形). $\because AD = 2DB$,$\therefore \frac{AD}{AB} = \frac{2}{3}$. $\because DF//BC$,$\therefore △ AFD ∽ △ ACB$,$\therefore \frac{AF}{AC} = \frac{AD}{AB} = \frac{2}{3}$,$\therefore \frac{S_{△ AFD}}{S_{△ ACB}} = (\frac{AD}{AB})^2 = \frac{4}{9}$,$\therefore$ 设$S_{△ AFD} = 4S$,$S_{△ ACB} = 9S$. $\because$ 沿DE将$△ ABC$剪成面积相等的两部分, $\therefore S_{△ ADE} = \frac{9}{2} S$,$\therefore \frac{S_{△ AFD}}{S_{△ ADE}} = \frac{4S}{\frac{9}{2} S} = \frac{8}{9} = \frac{AF}{AE}$,$\therefore \frac{AF}{AE} = \frac{AF}{AC} · \frac{AC}{AE} = \frac{AF}{AC} · \frac{AC}{AF} = \frac{8}{9} ÷ \frac{2}{3} = \frac{4}{3}$,$\therefore \frac{AE}{EC} = 3$.

2. 求面积比 [2025 合肥蜀山区期中] 如图，在$△ ABC$中，点$D,F$在边$AB$上，点$E,G$在边$AC$上，$DE// FG// BC$，如果$AD:DF:FB=1:2:3$，那么$S_{△ ADE}:S_{梯形DFGE}:S_{梯形FBCG}=$ 。

2 1:8:27 $\because AD: DF: FB = 1: 2: 3$,$\therefore AD: AF = 1: 3$, $AD: AB = 1: 6$. $\because DE//FG//BC$,$\therefore △ ADE ∽ △ AFG$,$△ ADE ∽ △ ABC$,$\therefore \frac{S_{△ ADE}}{S_{△ AFG}} = (\frac{AD}{AF})^2 = \frac{1}{9}$, $\frac{S_{△ ADE}}{S_{△ ABC}} = (\frac{AD}{AB})^2 = \frac{1}{36}$,$\therefore S_{△ ADE}: S_{梯形DFGE} = 1: 8$, $S_{△ ADE}: S_{梯形FBCG} = 1: 27$,$\therefore S_{△ ADE}: S_{梯形DFGE}: S_{梯形FBCG} = 1: 8: 27$.

3. 求线段长 如图，直角三角板ABC中，∠C = 90°，∠A = 30°，一边平行于BC的直尺将三角板ABC分成面积相等的三部分，若BC = √3，则EF的长为。

3 $2\sqrt{2} -2$ 由题意可知$EH//FG//BC$,$\therefore △ AFG ∽ △ ABC$,$△ AEH ∽ △ ABC$,$\therefore \frac{S_{△ AFG}}{S_{△ ABC}} = (\frac{FG}{BC})^2 = \frac{2}{3}$,$\frac{S_{△ AEH}}{S_{△ ABC}} = (\frac{EH}{BC})^2 = \frac{1}{3}$, $\therefore \frac{FG}{BC} = \frac{\sqrt{6}}{3}$,$\frac{EH}{BC} = \frac{\sqrt{3}}{3}$,$\therefore FG = \frac{\sqrt{6}}{3} × \sqrt{3} = \sqrt{2}$,$EH = \frac{\sqrt{3}}{3} × \sqrt{3} = 1$. 在Rt$△ AEH$中,$∠ A = 30°$,$\therefore AE = 2EH = 2$. 同理可得$AF = 2\sqrt{2}$,$\therefore EF = AF - AE = 2\sqrt{2} - 2$.