《2025年名师帮同步学案九年级数学全一册人教版》

1. 已知$\triangle ABC \backsim \triangle A^{\prime} B^{\prime} C^{\prime}$, 且$\frac{A B}{A^{\prime} B^{\prime}}=\frac{1}{2}$, 则$S_{\triangle A B C}: S_{\triangle A^{\prime} B^{\prime} C^{\prime}}$为(C)
A.$1: 2$
B.$2: 1$
C.$1: 4$
D.$4: 1$

答案： C

2. 如图, 直线$A D // B E // C F, B C= \frac{1}{2} A B, D E= 4$, 那么$E F$的值是(B)

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

答案： B

3.如图,在 Rt△ABC 中，∠ACB=90°,CD⊥AB,垂足为D.如果$\frac {C_{△ADC}} {C_{△CDB}}=\frac 32$，AD=9,那么BC 的长是(C)
A. 4 B. 6 C. $2 \sqrt{13}$ D. $3 \sqrt{10}$

答案： C 解析：$\because CD\perp AB,\therefore \angle ADC=\angle BDC=90^{\circ },\therefore \angle B+\angle BCD=90^{\circ }.\because \angle ACB=90^{\circ },\therefore \angle B+\angle A=90^{\circ },\therefore \angle BCD=\angle A,\therefore \triangle ADC\backsim \triangle CDB,\therefore \frac {C_{\triangle ADC}}{C_{\triangle CDB}}=\frac {AD}{CD}=\frac {AC}{BC}=\frac {3}{2}.\because AD=9,\therefore CD=6$，在$Rt\triangle ADC$中，$AC=\sqrt {AD^{2}+CD^{2}}=3\sqrt {13},\therefore BC=2\sqrt {13}$，故选 C.

4. 如图,$\triangle A B C$是等边三角形, 被一矩形所截,$A B$被截成三等分,$E H // B C$, 则四边形$E F G H的面积是\triangle A B C$的面积的(C)

A.$\frac{1}{9}$
B.$\frac{4}{9}$
C.$\frac{1}{3}$
D.$\frac{5}{9}$

答案： C 解析：$\because AB$被截为三等分，$\therefore \triangle AEH\backsim \triangle AFG\backsim \triangle ABC,\therefore \frac {AE}{AF}=\frac {1}{2},\frac {AE}{AB}=\frac {1}{3},\therefore S_{\triangle AFG}:S_{\triangle ABC}=4:9,S_{\triangle AEH}:S_{\triangle ABC}=1:9,\therefore S_{阴影}=\frac {4}{9}S_{\triangle ABC}-\frac {1}{9}S_{\triangle ABC}=\frac {1}{3}S_{\triangle ABC}$. 故选 C.

5. 如图所示是一个测试距离为$5 \mathrm{~m}$的视力表,根据这个视力表, 小华想制作一个测试距离为$3 \mathrm{~m}$的视力表, 则图中的$\frac{b_{2}}{b_{1}}= $$\frac {3}{5}$.

答案： $\frac {3}{5}$

6. 如图, 在$\triangle A B C$中,$D为A C$边上的点,$\angle D B C= \angle A, B C= \sqrt{6}, A C= 3$, 则$C D$的长为2.

答案： 2

7. 将一副三角尺按照如图所示的方式叠放在一起$(\angle B= 45^{\circ}, \angle D= 30^{\circ})$, 点$E是B C与A D$的交点, 则$\frac{D E}{A E}$的值为____$\sqrt {3}$.

答案： $\sqrt {3}$

8. 如图, 正方形$O A B C与正方形O D E F$是位似图形, 点$O$为位似中心, 相似比为$1: \sqrt{2}$, 点$A的坐标为(1,0)$, 则点$E$的坐标为____$(\sqrt {2},\sqrt {2})$.

答案： $(\sqrt {2},\sqrt {2})$

9. 如图, 已知$A B // C D, A D, B C相交于点E$, 点$F在E D$上, 且$\angle C B F= \angle D$.
(1) 求证:$F B^{2}= F E \cdot F A$;
证明：$\because AB// CD,$
$\therefore \angle A=\angle D.$
又$\because \angle CBF=\angle D,\therefore \angle A= \angle CBF$,
$\because \angle BFE= \angle AFB$,
$\therefore \triangle FBE \backsim \triangle FAB$,
$\therefore \frac {FB}{FA}=\frac {FE}{FB},\therefore FB^{2}=FE\cdot FA;$
(2) 若$B F= 3, E F= 2$, 求$\triangle A B E与\triangle B E F$的面积之比.
解：$\because FB^{2}=FE\cdot FA,BF=3,EF=2,$
$\therefore 3^{2}=2×(2+AE),$
$\therefore AE= \frac {5}{2},\therefore \frac {AE}{EF}= \frac {5}{4},$
$\therefore \triangle ABE$与$\triangle BEF$的面积之比为$5:4$.

答案：
(1) 证明：$\because AB// CD,$
$\therefore \angle A=\angle D.$
又$\because \angle CBF=\angle D,\therefore \angle A=\angle CBF,$
$\because \angle BFE=\angle AFB,$
$\therefore \triangle FBE\backsim \triangle FAB,$
$\therefore \frac {FB}{FA}=\frac {FE}{FB},\therefore FB^{2}=FE\cdot FA;$
(2) 解：$\because FB^{2}=FE\cdot FA,BF=3,EF=2,$
$\therefore 3^{2}=2×(2+AE),$
$\therefore AE=\frac {5}{2},\therefore \frac {AE}{EF}=\frac {5}{4},$
$\therefore \triangle ABE$与$\triangle BEF$的面积之比为$5:4$.