《2025年课时夺冠九年级数学上册湘教版》

1. (长沙天心区月考)已知$\triangle ABC \backsim \triangle A'B'C'$,$\frac { A B } { A ^ { \prime } B ^ { \prime } } = \frac { 3 } { 4 }$,则$\triangle ABC与\triangle A ^ { \prime } B ^ { \prime } C ^ { \prime }$的面积之比为()
A. $\frac { 3 } { 4 }$
B. $\frac { 4 } { 3 }$
C. $\frac { 9 } { 16 }$
D. $\frac { 16 } { 9 }$

答案： C

2. 已知$\triangle ABC \backsim \triangle DEF$,相似比为2,且$\triangle ABC$的面积为16,则$\triangle DEF$的面积为()
A. 32
B. 8
C. 4
D. 16

答案： C

3. 如图,点D,E分别在$\triangle ABC$的边AC,AB上,$\triangle ADE \backsim \triangle ABC$,M,N分别是DE,BC的中点. 若$\frac { A M } { A N } = \frac { 1 } { 2 }$,则$\frac { S _ { \triangle A D E } } { S _ { \triangle A B C } }$的值是______.

答案： $\frac{1}{4}$

4. 如图,四边形ABCD为平行四边形,E,F为边CD的两个三等分点,连接AF,BE交于点G,则$S _ { \triangle E F G } : S _ { \triangle A B G } = $______.

答案： $1:9$

5. 如图,在$\triangle ABC$中,$DE // BC$,且$S _ { \triangle A D E } : S _ { 四边形 BCED } = 1 : 2$,$BC = 2 \sqrt { 6 }$,求DE的长.

答案： 解：$\because DE// BC,\therefore \triangle ADE\backsim \triangle ABC,$
$\therefore \frac{S_{\triangle ADE}}{S_{\triangle ABC}} = (\frac{DE}{BC})^2.\because \frac{S_{\triangle ADE}}{S_{四边形BCED}} = \frac{1}{2},\therefore \frac{S_{\triangle ADE}}{S_{\triangle ABC}} = \frac{1}{3},\therefore \frac{DE}{BC} = \frac{1}{\sqrt{3}}$，即$\frac{DE}{2\sqrt{6}} = \frac{1}{\sqrt{3}}$，
$\therefore DE = 2\sqrt{2}$.

6. 若$\triangle ADE \backsim \triangle ABC$,且$AD : AB = 1 : 2$,则$\triangle ADE与\triangle ABC$的周长之比是()
A. $1 : 2$
B. $1 : 3$
C. $2 : 1$
D. $1 : 4$

答案： A

7. 如果$\triangle ABC \backsim \triangle DEF$,且$\triangle ABC$的三边长分别为3,5,6,$\triangle DEF$的最短边长为9,那么$\triangle DEF$的周长为()
A. 14
B. $\frac { 126 } { 5 }$
C. 21
D. 42

答案： D

8. 如图,在$\triangle ABC$中,ED交AB于点E,交AC于点D,$\frac { A D } { A B } = \frac { A E } { A C } = \frac { 3 } { 5 }$,且$\triangle ABC的周长与\triangle ADE$的周长差是16,求$\triangle ABC和\triangle ADE$的周长.

答案： 解：$\because \angle A = \angle A,\frac{AD}{AB} = \frac{AE}{AC} = \frac{3}{5},\therefore \triangle ADE\backsim$
$\triangle ABC$，且相似比为$\frac{3}{5}$. 设$\triangle ABC$的周长为$5x$，则$\triangle ADE$的周长为$3x$. 由题意，得$5x - 3x = 16$，解得$x = 8$，$\therefore \triangle ABC$的周长为$40$，$\triangle ADE$的周长为$24$.

9. 如图,在$Rt \triangle ABC$中,$\angle A C B = 90 ^ { \circ }$,$\angle A = 30 ^ { \circ }$,$CD \perp AB$于点D. 求$\triangle BCD与\triangle ABC$的周长之比.

答案： 解：$\because CD\perp AB,\therefore \angle BDC = 90^\circ = \angle ACB.\because \angle B = \angle B,\therefore \triangle BCD\backsim$
$\triangle BAC,\therefore \angle BCD = \angle A = 30^\circ,\therefore BC = 2BD.\because \triangle BCD\backsim \triangle BAC,\therefore C_{\triangle BCD}:$
$C_{\triangle ABC} = BD:BC = 1:2$.