1. (2024·内江)已知△ABC与△DEF相似,且相似比为$\frac{1}{3}$,则△ABC与△DEF的周长之比是( )
A. 1:1
B. 1:3
C. 1:6
D. 1:9
A. 1:1
B. 1:3
C. 1:6
D. 1:9
答案
B
2. 如图,在矩形ABCD中,E是边AD上一点,且AE = 2DE,BD与CE相交于点F. 若△DEF的面积是3,则△BCF的面积是( )

A. 9
B. 12
C. 27
D. 48
A. 9
B. 12
C. 27
D. 48
答案
C
3. (2023·泰州)两个相似图形的周长比为3:2,则它们的面积比为__________.
答案
9 : 4
4. 一块3 m×2 m的矩形广告牌的制作成本是120元,在每平方米制作成本相同的情况下,若将广告牌的四边都扩大为原来的3倍,则扩大后一块矩形广告牌的制作成本是__________元.
答案
1080
5. 如图,在△ABC和△DEC中,∠A = ∠D,∠BCE = ∠ACD.
(1)求证:△ABC∽△DEC;
(2)若$S_{\triangle ABC}:S_{\triangle DEC}=4:9$,BC = 6,求EC的长.

(1)求证:△ABC∽△DEC;
(2)若$S_{\triangle ABC}:S_{\triangle DEC}=4:9$,BC = 6,求EC的长.
答案
(1) $\because \angle BCE = \angle ACD, \therefore \angle BCE + \angle ACE = \angle ACD + \angle ACE. \therefore \angle ACB = \angle DCE$. 又 $\because \angle A = \angle D, \therefore \triangle ABC \backsim \triangle DEC$
(2) $\because \triangle ABC \backsim \triangle DEC, \therefore \frac{S_{\triangle ABC}}{S_{\triangle DEC}} = (\frac{BC}{EC})^2$.
$\because S_{\triangle ABC} : S_{\triangle DEC} = 4 : 9, \therefore \frac{BC}{EC} = \frac{2}{3}. \because BC = 6, \therefore EC = 9$
(2) $\because \triangle ABC \backsim \triangle DEC, \therefore \frac{S_{\triangle ABC}}{S_{\triangle DEC}} = (\frac{BC}{EC})^2$.
$\because S_{\triangle ABC} : S_{\triangle DEC} = 4 : 9, \therefore \frac{BC}{EC} = \frac{2}{3}. \because BC = 6, \therefore EC = 9$
6. 两个相似三角形的最短边的长分别为5 cm和3 cm,他们的周长之差为12 cm,那么大三角形的周长为( )
A. 14 cm
B. 16 cm
C. 18 cm
D. 30 cm
A. 14 cm
B. 16 cm
C. 18 cm
D. 30 cm
答案
D
7. 如图,在△ABC中,AC = 2,BC = 4,D为边BC上的一点,且∠CAD = ∠B. 若△ADC的面积为a,则△ABD的面积为( )

A. 2a
B. $\frac{5}{2}a$
C. 3a
D. $\frac{7}{2}a$
A. 2a
B. $\frac{5}{2}a$
C. 3a
D. $\frac{7}{2}a$
答案
C
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