《2025年通成学典课时作业本八年级数学上册苏科版宿迁专版》

15. 已知直角三角形纸片的两条直角边的长分别为m和n(m＜n),过锐角顶点把该纸片剪成两个三角形.若这两个三角形都为等腰三角形,则下列结论正确的是 ( )

A.$m^2+2mn+n^2= 0$
B.$m^2-2mn+n^2= 0$
C.$m^2+2mn-n^2= 0$
D.$m^2-2mn-n^2= 0$

答案：
C 解析:如图,剪成一个腰长为m的等腰直角三角形和一个腰长为n - m的等腰三角形.由题意,得$m^{2}+m^{2}=(n - m)^{2}$,即$2m^{2}=n^{2}-2mn+m^{2}$,
∴$m^{2}+2mn - n^{2}=0$.

16. (整体思想)如图,C是线段AB上的一点,分别以AC,BC为边向两侧作正方形,设AB= 6,两个正方形$S_1,S_2$的面积之和为20,则△BCD的面积为______.

答案： 4 解析:设AC = a,BC = b.由题意,得a + b = 6,$a^{2}+b^{2}=20$.
∵$a^{2}+b^{2}=(a + b)^{2}-2ab$,
∴$20=6^{2}-2ab$,
∴ab = 8,
∴△BCD的面积为$\frac{1}{2}ab=\frac{1}{2}×8=4$.

17. (2023·广州)如图,正方形ABCD的边长为4,点E边BC上,且BE= 1,F为对角线BD上一动点,连接CF,EF,则CF+EF的最小值为______.

答案： $\sqrt{17}$

18. (方程思想)如图,AC平分∠BAD,AD= BC= CD= 3,AC= 5,则AB长为______.

答案： $\frac{16}{3}$ 解析:在AB上截取AE = AD,连接CE,过点C作CF⊥AB于点F.
∵AC平分∠BAD,
∴∠DAC = ∠EAC.又
∵AD = AE,AC = AC,
∴△DAC≌△EAC,
∴CD = CE.
∵AD = BC = CD = 3,
∴CE = BC = 3.
∵CF⊥AB,
∴EF = BF.设EF = x.在Rt△AFC中,AE = AD = 3,AC = 5,
∴AF = x + 3,
∴$CF^{2}=5^{2}-(x + 3)^{2}$.在Rt△ECF中,由勾股定理,得$CF^{2}=3^{2}-x^{2}$.
∴$5^{2}-(x + 3)^{2}=3^{2}-x^{2}$,解得$x=\frac{7}{6}$,
∴$EF=\frac{7}{6}$,
∴$AB=AE + 2EF=3 + 2×\frac{7}{6}=\frac{16}{3}$.

19..如图,在△ABC中,AB= 8,AC= 17,AD是边BC上中线,AD= $\frac{15}{2}$,求△ABC的面积.

答案： 延长AD至点E,使得ED = AD,连接CE.
∵AD是边BC上的中线,
∴BD = CD.在△ABD和△ECD中,$\left\{\begin{array}{l} AD = ED,\\ ∠ADB = ∠EDC,\\ BD = CD,\end{array}\right.$
∴△ABD≌△ECD,
∴AB = EC = 8,$S_{\triangle ABD}=S_{\triangle ECD}$.
∵$S_{\triangle ABC}=S_{\triangle ABD}+S_{\triangle ADC}$,$S_{\triangle AEC}=S_{\triangle ECD}+S_{\triangle ADC}$,
∴$S_{\triangle ABC}=S_{\triangle AEC}$.
∵$AD=\frac{15}{2}$,
∴AE = AD + DE = 15.又
∵AC = 17,
∴$AE^{2}+EC^{2}=AC^{2}$,
∴∠E = 90°,
∴$S_{\triangle ABC}=S_{\triangle AEC}=\frac{1}{2}AE\cdot EC=\frac{1}{2}×15×8=60$.

20. 如图,在一张长方形纸片ABCD中,AB= 8,BC= 6,P为AD上的一点,将△ABP沿BP折叠至△EBP处,PE与CD相交于点O,且OE= OD,求AP的长.

答案： 设DC与BE相交于点G.
∵四边形ABCD是长方形,
∴∠D = ∠A = ∠C = 90°,AD = BC = 6,CD = AB = 8.根据折叠的特征,得△EBP≌△ABP,
∴EP = AP,BE = BA = 8,∠E = ∠A = 90°,
∴∠D = ∠E.在△ODP和△OEG中,$\left\{\begin{array}{l} ∠D = ∠E,\\ OD = OE,\\ ∠DOP = ∠EOG,\end{array}\right.$
∴△ODP≌△OEG,
∴OP = OG,PD = GE.
∴OP + OE = OG + OD,即EP = DG.设AP = x,则PD = GE = 6 - x,EP = DG = x.
∴CG = 8 - x,BG = 8 - (6 - x)=2 + x.在Rt△BCG中,由勾股定理,得$BC^{2}+CG^{2}=BG^{2}$,即$6^{2}+(8 - x)^{2}=(2 + x)^{2}$,解得x = 4.8.
∴AP的长为4.8.