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1.(2023·湖南中考)下列长度的各组线段能组成一个三角形的是()
A. 1 cm,2 cm,3 cm
B. 3 cm,8 cm,5 cm
C. 4 cm,5 cm,10 cm
D. 4 cm,5 cm,6 cm
A. 1 cm,2 cm,3 cm
B. 3 cm,8 cm,5 cm
C. 4 cm,5 cm,10 cm
D. 4 cm,5 cm,6 cm
答案
D
2. 若一个等腰三角形两内角的度数之比为1:4,则这个等腰三角形的底角的度数为()
A. $30^{\circ}$
B. $30^{\circ}或120^{\circ}$
C. $120^{\circ}$
D. $30^{\circ}或80^{\circ}$
A. $30^{\circ}$
B. $30^{\circ}或120^{\circ}$
C. $120^{\circ}$
D. $30^{\circ}或80^{\circ}$
答案
D
3.(2025·天水期中)比较2,$\sqrt{5}$,$\sqrt[3]{7}$的大小,正确的是()
A. $2＜\sqrt{5}＜\sqrt[3]{7}$
B. $2＜\sqrt[3]{7}＜\sqrt{5}$
C. $\sqrt{5}＜\sqrt[3]{7}＜2$
D. $\sqrt[3]{7}＜2＜\sqrt{5}$
A. $2＜\sqrt{5}＜\sqrt[3]{7}$
B. $2＜\sqrt[3]{7}＜\sqrt{5}$
C. $\sqrt{5}＜\sqrt[3]{7}＜2$
D. $\sqrt[3]{7}＜2＜\sqrt{5}$
答案
D
4. 如图,在$\triangle ABC$中,$AB = AC = 5$,$BC = 6$,点$M为BC$边上的中点,$MN\perp AC于点N$,那么$MN$等于()
A. $\frac{6}{5}$
B. $\frac{8}{5}$
C. $\frac{12}{5}$
D. $\frac{24}{5}$
A. $\frac{6}{5}$
B. $\frac{8}{5}$
C. $\frac{12}{5}$
D. $\frac{24}{5}$
答案
B
5.(2024·宿迁校级月考)如图,在数轴上,点$A$,$B表示的数分别为-1$,1,$CB\perp AB于点B$,且$BC = 1$.连接$AC$,在$AC上截取CD = BC$,以点$A$为圆心,$AD$的长为半径画弧,交线段$AB于点E$,则点$E$表示的实数是()
A. $\sqrt{5}-1$
B. $\sqrt{5}-2$
C. $\sqrt{5}$
D. $1-\sqrt{5}$
A. $\sqrt{5}-1$
B. $\sqrt{5}-2$
C. $\sqrt{5}$
D. $1-\sqrt{5}$
答案
A
6.(2023·台州中考)如图,锐角三角形$ABC$中,$AB = AC$,点$D$,$E分别在边AB$,$AC$上,连接$BE$,$CD$.下列命题中,假命题是()
A. 若$CD = BE$,则$\angle DCB= \angle EBC$
B. 若$\angle DCB= \angle EBC$,则$CD = BE$
C. 若$BD = CE$,则$\angle DCB= \angle EBC$
D. 若$\angle DCB= \angle EBC$,则$BD = CE$
A. 若$CD = BE$,则$\angle DCB= \angle EBC$
B. 若$\angle DCB= \angle EBC$,则$CD = BE$
C. 若$BD = CE$,则$\angle DCB= \angle EBC$
D. 若$\angle DCB= \angle EBC$,则$BD = CE$
答案
A 解析:连接DF.∵在Rt△ABC中,∠ACB=90°,AC=3,BC=4,∴AB=5.∵AD=AC=3,AF⊥CD,∴CE=DE,BD=AB−AD=2,∴CF=DF.在△ADF和△ACF中,$\left\{\begin{array}{l} AD = AC,\\ DF = CF,\\ AF = AF,\end{array}\right.$∴△ADF≌△ACF(SSS),∴∠ADF=∠ACF=90°,∴∠BDF=90°.设CF=DF=x,则BF=4−x.在Rt△BDF中,由勾股定理,得$DF^{2}+BD^{2}=BF^{2}$,即$x^{2}+2^{2}=(4 - x)^{2}$,解得x=1.5,∴CF=1.5,故选A.
7. 如图,在$Rt\triangle ABC$中,$\angle ACB = 90^{\circ}$,$AC = 3$,$BC = 4$,点$D在AB$上,$AD = AC$,$AF\perp CD$,交$CD于点E$,交$CB于点F$,则$CF$的长是()
A. 1.5
B. 1.8
C. 2
D. 2.5
A. 1.5
B. 1.8
C. 2
D. 2.5
答案
A 解析:连接DF.∵在Rt△ABC中,∠ACB=90°,AC=3,BC=4,∴AB=5.∵AD=AC=3,AF⊥CD,∴CE=DE,BD=AB−AD=2,∴CF=DF.在△ADF和△ACF中,$\left\{\begin{array}{l} AD = AC,\\ DF = CF,\\ AF = AF,\end{array}\right.$∴△ADF≌△ACF(SSS),∴∠ADF=∠ACF=90°,∴∠BDF=90°.设CF=DF=x,则BF=4−x.在Rt△BDF中,由勾股定理,得$DF^{2}+BD^{2}=BF^{2}$,即$x^{2}+2^{2}=(4 - x)^{2}$,解得x=1.5,∴CF=1.5,故选A.
8. 如图,$\triangle ABC$中,$BC = 10$,$AC - AB = 4$.过点$C作\angle BAC$的平分线的垂线,垂足为点$D$,连接$BD$,则$S_{\triangle BDC}$的最大值为()
A. 10
B. 15
C. 12
D. 14
A. 10
B. 15
C. 12
D. 14
答案
A 解析:延长AB,CD交于点E.∵AD平分∠BAC,∴∠CAD=∠EAD.∵CD⊥AD,∴∠ADC=∠ADE=90°.在△ADE和△ADC中,$\left\{\begin{array}{l} ∠ADE = ∠ADC,\\ AD = AD,\\ ∠EAD = ∠CAD,\end{array}\right.$∴△ADE≌△ADC(ASA),∴AC=AE,DE=CD.∵AC−AB=4,∴AE−AB=4,即BE=4.∵DE=DC,∴$S_{△BDC}=\frac{1}{2}S_{△BEC}$,∴当BE⊥BC时,$S_{△BDC}$的面积最大,即△BDC的最大面积=$\frac{1}{2}×\frac{1}{2}×10×4 = 10$.故选A.
9. 已知等腰三角形的周长为7,一边长为3,则它的腰长为______.
答案
2或3
10. 如图,在$\triangle ABC$中,$AD\perp BC于点D$,要使$\triangle ABD\cong\triangle ACD$,若根据“HL”判定,还需要加条件______.
答案
AB=AC
11.(2023·丽水中考)如图,在$\triangle ABC$中,$AC的垂直平分线交BC于点D$,交$AC于点E$,$\angle B= \angle ADB$.若$AB = 4$,则$DC$的长是______.
答案
4
