登录
6. (2025·双流)在菱形$ABCD$中,若对角线$AC = 8cm$,$BD = 6cm$,则菱形$ABCD$的周长是()
A. $25cm$
B. $20cm$
C. $15cm$
D. $10cm$
A. $25cm$
B. $20cm$
C. $15cm$
D. $10cm$
答案
B
7. (2025·编写)如图,在菱形$ABCD$中,$\angle C = 80^{\circ}$,则$\angle ABD$的度数为()
A. $80^{\circ}$
B. $70^{\circ}$
C. $60^{\circ}$
D. $50^{\circ}$
A. $80^{\circ}$
B. $70^{\circ}$
C. $60^{\circ}$
D. $50^{\circ}$
答案
D
8. (2023·南京)如图,四边形$ABCD$是菱形,对角线$AC与BD相交于点O$,$DH\perp BC于点H$。若$AC = 8$,$BD = 6$,则$DH$的长度为()
A. $\frac{48}{5}$
B. $\frac{36}{5}$
C. $\frac{24}{5}$
D. $4$
A. $\frac{48}{5}$
B. $\frac{36}{5}$
C. $\frac{24}{5}$
D. $4$
答案
C
9. (1)(2022·天府新区)如图,四边形$ABCD$是菱形,$AE\perp CD于点E$,$AF\perp BC于点F$。求证：$CE = CF$。
(2)(2023·台州)如图,菱形$ABCD的对角线AC$,$BD相交于点O$,过点$D作DH\perp AB于点H$,连接$OH$。若$OA = 4$,$OH = 3$,求$DH$的长。
(2)(2023·台州)如图,菱形$ABCD的对角线AC$,$BD相交于点O$,过点$D作DH\perp AB于点H$,连接$OH$。若$OA = 4$,$OH = 3$,求$DH$的长。
答案
(1)【证明】∵四边形$ABCD$是菱形，∴$AB = BC = CD = AD$，$∠B = ∠D$．∵$AE⊥CD$，$AF⊥BC$，∴$∠AED = ∠AFB = 90^{\circ}$．在$\triangle ADE$和$\triangle ABF$中，$\left\{\begin{array}{l} ∠AED = ∠AFB,\\ ∠D = ∠B,\\ AD = AB,\end{array}\right.$ ∴$\triangle ADE≌\triangle ABF(AAS)$，∴$DE = BF$，∴$CD - DE = BC - BF$，∴$CE = CF$．
(2)【解】∵四边形$ABCD$是菱形，∴$OC = OA = 4$，$OB = OD$，$AC⊥BD$，∴$AC = 2OA = 8$．∵$DH⊥AB$，∴$∠BHD = 90^{\circ}$，∴$BD = 2OH$．∵$OH = 3$，∴$BD = 6$，∴菱形$ABCD$的面积$=\frac{1}{2}AC\cdot BD=\frac{1}{2}\times8\times6 = 24$，$AB = \sqrt{OA^{2} + OB^{2}} = \sqrt{4^{2} + 3^{2}} = 5$．∵菱形$ABCD$的面积$= AB\cdot DH = 5DH = 24$，∴$DH = \frac{24}{5}$．
(2)【解】∵四边形$ABCD$是菱形，∴$OC = OA = 4$，$OB = OD$，$AC⊥BD$，∴$AC = 2OA = 8$．∵$DH⊥AB$，∴$∠BHD = 90^{\circ}$，∴$BD = 2OH$．∵$OH = 3$，∴$BD = 6$，∴菱形$ABCD$的面积$=\frac{1}{2}AC\cdot BD=\frac{1}{2}\times8\times6 = 24$，$AB = \sqrt{OA^{2} + OB^{2}} = \sqrt{4^{2} + 3^{2}} = 5$．∵菱形$ABCD$的面积$= AB\cdot DH = 5DH = 24$，∴$DH = \frac{24}{5}$．
10. (2023·湖北)如图,四边形$ABCD$是菱形,$AE\perp BC于点E$,$AF\perp CD于点F$。
(1)求证：$\triangle ABE\cong\triangle ADF$；
(2)若$AE = 4$,$CF = 2$,求菱形的边长。
(1)求证：$\triangle ABE\cong\triangle ADF$；
(2)若$AE = 4$,$CF = 2$,求菱形的边长。
答案
(1)【证明】∵四边形$ABCD$是菱形，∴$AB = BC = CD = AD$，$∠B = ∠D$．∵$AE⊥BC$，$AF⊥CD$，∴$∠AEB = ∠AFD = 90^{\circ}$．在$\triangle ABE$和$\triangle ADF$中，$\left\{\begin{array}{l} ∠AEB = ∠AFD,\\ ∠B = ∠D,\\ AB = AD,\end{array}\right.$ ∴$\triangle ABE≌\triangle ADF(AAS)$．
(2)【解】设菱形的边长为$x$．∵$AB = CD = x$，$CF = 2$，∴$DF = x - 2$．∵$\triangle ABE≌\triangle ADF$，∴$BE = DF = x - 2$．
(2)【解】设菱形的边长为$x$．∵$AB = CD = x$，$CF = 2$，∴$DF = x - 2$．∵$\triangle ABE≌\triangle ADF$，∴$BE = DF = x - 2$．
11. (1)(2023·武侯)如图,菱形$ABCD的对角线交于原点O$,若点$A的坐标为(-3,-5)$,点$B的坐标为(10,m)$,点$D的坐标为(n,6)$,则边$CD = $
______。
(2)(2025·编写)如图,在菱形$ABCD$中,$E$,$F分别是边CD$,$BC$上的动点,连接$AE$,$EF$,$G$,$H分别为AE$,$EF$的中点,连接$GH$。若$\angle B = 45^{\circ}$,$BC = 2\sqrt{10}$,则$GH$的最小值为______。
______。
(2)(2025·编写)如图,在菱形$ABCD$中,$E$,$F分别是边CD$,$BC$上的动点,连接$AE$,$EF$,$G$,$H分别为AE$,$EF$的中点,连接$GH$。若$\angle B = 45^{\circ}$,$BC = 2\sqrt{10}$,则$GH$的最小值为______。
答案
(1)$\sqrt{170}$ (2)$\sqrt{5}$
12. (2023·丽水)如图,在菱形$ABCD$中,$AD = 10$,$AC = 12$,点$E是点A关于直线CD$的对称点,连接$AE交CD于点F$,连接$CE$,$DE$,则$AE$的长是______。
答案
19.2
