《2025年轻松作业本八年级数学上册苏科版》

9. 一个三角形的三边长分别为$2$、$5$、$x$,另一个三角形的三边长分别为$y$、$2$、$6$,若这两个三角形全等,则$x + y$的值为(A)
A. $11$
B. $7$
C. $8$
D. $13$

答案： A

10. 如图,$\triangle ABC \cong \triangle ADE$。
(1) 若$\triangle ABC周长为24$,$AD = 6$,$AE = 9$,则$BC = $9；
(2) 若$\angle BAD = 42^{\circ}$,则$\angle EFC = $42$^{\circ}$。

答案：
(1) 9
(2) 42

11. 如图,$AB$、$CD相交于点E$,若$\triangle ABC \cong \triangle ADE$,$\angle BAC = 28^{\circ}$,则$\angle B$的度数是______48$^{\circ}$。

答案： 48

12. 如图,已知$\triangle EAB \cong \triangle DCE$,$AB$、$EC$分别是两个三角形的最长边,$\angle A = \angle C = 35^{\circ}$,$\angle CDE = 100^{\circ}$,$\angle DEB = 10^{\circ}$,求$\angle AEC$的度数。
解：$\because \triangle EAB \cong \triangle DCE$，$\therefore \angle BEA = \angle CDE = 100^{\circ}$。$\because \angle A = \angle C = 35^{\circ}$，$\angle CDE = 100^{\circ}$，$\therefore \angle DEC = 180^{\circ} - 100^{\circ} - 35^{\circ} = 45^{\circ}$。$\because \angle DEB = 10^{\circ}$，$\therefore \angle BEC = 45^{\circ} - 10^{\circ} = 35^{\circ}$。$\therefore \angle CEA = 100^{\circ} - 35^{\circ} = $65°。

答案： 解：$\because \triangle EAB \cong \triangle DCE$，$\therefore \angle BEA = \angle CDE = 100^{\circ}$。$\because \angle A = \angle C = 35^{\circ}$，$\angle CDE = 100^{\circ}$，$\therefore \angle DEC = 180^{\circ} - 100^{\circ} - 35^{\circ} = 45^{\circ}$。$\because \angle DEB = 10^{\circ}$，$\therefore \angle BEC = 45^{\circ} - 10^{\circ} = 35^{\circ}$。$\therefore \angle CEA = 100^{\circ} - 35^{\circ} = 65^{\circ}$。

13. 如图,已知$\triangle ABD \cong \triangle CFD$,$AD \perp BC于D$。
(1) 求证：$CE \perp AB$；
证明：$\because \triangle ABD \cong \triangle CFD$，$\therefore \angle BAD = \angle DCF$。又 $\because \angle AFE = \angle CFD$，$\therefore \angle AEF = \angle CDF = 90^{\circ}$。$\therefore CE \perp AB$；
(2) 已知$BC = 7$,$AD = 5$,求$AF$的长。
解：$\because \triangle ABD \cong \triangle CFD$，$\therefore BD = DF$，$AD = CD$。$\because BC = 7$，$AD = DC = 5$，$\therefore BD = BC - CD = 2$。$\therefore AF = AD - DF = 5 - 2 =$3。

答案：
(1) 证明：$\because \triangle ABD \cong \triangle CFD$，$\therefore \angle BAD = \angle DCF$。又 $\because \angle AFE = \angle CFD$，$\therefore \angle AEF = \angle CDF = 90^{\circ}$。$\therefore CE \perp AB$；
(2) 解：$\because \triangle ABD \cong \triangle CFD$，$\therefore BD = DF$，$AD = CD$。$\because BC = 7$，$AD = DC = 5$，$\therefore BD = BC - CD = 2$。$\therefore AF = AD - DF = 5 - 2 = 3$。

14. 如图,已知$\triangle ABC \cong \triangle DEB$,点$E在AB$上,$DE与AC相交于点F$。
(1) 当$DE = 8$,$BC = 5$时,求线段$AE$的长；3
(2) 已知$\angle D = 35^{\circ}$,$\angle C = 60^{\circ}$,求$\angle DBC与\angle AFD$的度数。25°，130°

答案： 解：
(1) $\because \triangle ABC \cong \triangle DEB$，$DE = 8$，$BC = 5$，$\therefore AB = DE = 8$，$BE = BC = 5$。$\therefore AE = AB - BE = 8 - 5 = 3$；
(2) $\because \triangle ABC \cong \triangle DEB$，$\angle D = 35^{\circ}$，$\angle C = 60^{\circ}$，$\therefore \angle DBE = \angle C = 60^{\circ}$，$\angle A = \angle D = 35^{\circ}$，$\angle ABC = \angle DEB$。$\therefore \angle ABC = 180^{\circ} - \angle A - \angle C = 85^{\circ}$。$\therefore \angle DBC = \angle ABC - \angle DBE = 85^{\circ} - 60^{\circ} = 25^{\circ}$。$\because \angle ABC = 85^{\circ}$，$\therefore \angle DEB = 85^{\circ}$。$\therefore \angle AED = 95^{\circ}$。$\therefore \angle AFD = \angle A + \angle AED = 35^{\circ} + 95^{\circ} = 130^{\circ}$。