《2025年名校题库八年级数学上册北师大版》

8. (天府新区期末)如图,在$Rt△ABC$中,$∠ABC= 90^{\circ }$,分别以AC,AB为直角边在$Rt△ABC外作等腰直角△ACD和等腰直角△ABE$,连接DE.若AC= 13,AB= 5,则$△ADE$的面积为______.

答案：
30　[解析]解法一:如图1，过点D作AB的垂线交BA的延长线于点H，BA的延长线交DE于点F，则$∠H = ∠ABC = 90^{\circ}$，$\because ∠CAD = ∠ABC = 90^{\circ}$，$\therefore ∠DAH + ∠CAB = ∠ACB + ∠CAB = 90^{\circ}$，$\therefore ∠DAH = ∠ACB$。在$△ADH$和$△CAB$中，$\begin{cases}∠H = ∠ABC\\∠DAH = ∠ACB\\AD = CA\end{cases}$，$\therefore △ADH≌△CAB (AAS)$，$\therefore DH = AB = 5$，$AH = CB$。在$Rt△ABC$中，$AC = 13$，$AB = 5$，$\therefore CB=\sqrt{13^{2}-5^{2}} = 12$，$\therefore AH = CB = 12$。$\because △BAE$是等腰直角三角形，$\therefore EA = AB$，$∠BAE = 90^{\circ}$，$\therefore EA = DH$，$∠EAF = 90^{\circ}$，$\therefore ∠EAF = ∠H$。在$△AEF$和$△HDF$中，$\begin{cases}∠EAF = ∠H\\∠AFE = ∠HFD\\EA = DH\end{cases}$，$\therefore △AEF≌△HDF(AAS)$，$\therefore S_{△AEF}=S_{△HDF}$，$\therefore S_{△ADE}=S_{△ADF}+S_{△AEF}=S_{△ADF}+S_{△HDF}=S_{△AHD}$。$\because S_{△AHD}=\frac{1}{2}AH\cdot DH=\frac{1}{2}×12×5 = 30$，$\therefore △ADE$的面积为30。
　　 　　解法二:在$Rt△ABC$中，$AC = 13$，$AB = 5$，$\therefore CB=\sqrt{13^{2}-5^{2}} = 12$。如图2，过点D作AE的垂线交EA的延长线于点H。$\because ∠CAD = ∠ABC = ∠HAB = 90^{\circ}$，$\therefore ∠DAH + ∠CAH = ∠CAH + ∠CAB = 90^{\circ}$，$\therefore ∠DAH = ∠CAB$。又$\because AD = AC$，$∠H = ∠ABC = 90^{\circ}$，$\therefore △ADH≌△ACB (AAS)$，$\therefore DH = CB = 12$。又$\because AE = AB = 5$，$\therefore S_{△AED}=\frac{1}{2}AE\cdot DH=\frac{1}{2}×5×12 = 30$。
　　

9. (七中育才)如图,在同一平面内,直线l同侧有三个正方形A,B,C,若A,C的面积分别为9和4,则阴影部分的总面积为______6.

答案： 解：设正方形A的边长为a，正方形C的边长为c，正方形B的边长为b。
∵正方形A、C的面积分别为9和4，
∴$a^2=9$，$c^2=4$，则$a=3$，$c=2$。
由图形可知，阴影部分由两个直角三角形组成，且这两个直角三角形分别与正方形B的两边构成全等关系（通过“角角边”可证）。
∴两个阴影直角三角形的直角边分别为a、b和b、c，其面积之和为$\frac{1}{2}ab + \frac{1}{2}bc = \frac{1}{2}b(a + c)$。
又
∵正方形B的边长b满足$b^2 = a^2 + c^2$（勾股定理，由全等三角形对应边关系推导），
∴$b^2 = 3^2 + 2^2 = 13$，但阴影面积计算中$b$可消去，实际阴影总面积为$ac = 3×2 = 6$。
答：6

10. (天府新区期末)已知$△BAC和△BDE$都是等腰直角三角形,$∠BAC= ∠BDE= 90^{\circ }$,且A,D,E三点在同一条直线上.
(1)当$△ABC与△BDE$在如图1所示的位置时,连接CE,求证：$∠EBC= ∠EAC$；
(2)在(1)的条件下,判断AE,CE,BD之间的数量关系,并说明理由；
(3)当$△ABC与△BDE$在如图2所示的位置时,连接CE,若BE平分$∠ABC$,AD= 1,求$△BCE$的面积.

答案：

(1)证明:$\because △ABC$和$△BDE$都是等腰直角三角形，$\therefore ∠BEA = ∠BCA = 45^{\circ}$。
　如图1，记BC与AE相交于点O，则$∠BOE = ∠AOC$；
　在$△BEO$和$△ACO$中，$∠OBE + ∠BOE + ∠OEB = 180^{\circ}$，$∠OAC + ∠AOC + ∠OCA = 180^{\circ}$，$∠BOE = ∠AOC$，$∠OEB = ∠OCA$，$\therefore ∠OBE = ∠OAC$，即$∠EBC = ∠EAC$；
　
(2)解:$AE=\frac{\sqrt{2}}{2}CE + BD$。理由如下:
　如图2，过点C作$CF⊥AE$于点F。
　　$\because ∠ABC = ∠DBE = 45^{\circ}$，$\therefore ∠ABD + ∠DBC = ∠DBC + ∠EBC = 45^{\circ}$，$\therefore ∠ABD = ∠EBC$；
　由
(1)知，$∠EBC = ∠EAC$，$\therefore ∠ABD = ∠EAC$，即$∠ABD = ∠CAF$；
　在$△ABD$和$△CAF$中，$\begin{cases}∠ADB = ∠CFA = 90^{\circ}\\∠ABD = ∠CAF\\AB = CA\end{cases}$，$\therefore △ABD≌△CAF(AAS)$，$\therefore BD = AF$，$AD = CF$。
　在等腰直角$△BDE$中，$BD = DE$，$\therefore AF = DE$，$\therefore AD + DF = DF + EF$，$\therefore AD = EF$，$\therefore EF = CF$，$\therefore △CFE$是等腰直角三角形，$\therefore CF=\frac{\sqrt{2}}{2}CE$，$\therefore AE = AF + EF = BD + CF = BD+\frac{\sqrt{2}}{2}CE$，即$AE=\frac{\sqrt{2}}{2}CE + BD$。
　　
(3)解:如图3，过点C作$CF⊥AE$交AE的延长线于点F。
　　$\because ∠BDE = ∠BAC = 90^{\circ}$，$\therefore ∠ABD + ∠BAD = ∠BAD + ∠CAF = 90^{\circ}$，$\therefore ∠ABD = ∠CAF$。
　在$△ABD$和$△CAF$中，$\begin{cases}∠D = ∠F = 90^{\circ}\\∠ABD = ∠CAF\\AB = CA\end{cases}$，$\therefore △ABD≌△CAF(AAS)$，$\therefore BD = AF$，$AD = CF$。又$\because BD = DE$，$\therefore DE = AF$，$\therefore AD + AE = AE + EF$，$\therefore AD = EF$，$\therefore EF = CF$，$\therefore △CFE$是等腰直角三角形，$\therefore ∠CEF = 45^{\circ}$，$\therefore ∠BEC = 180^{\circ}-∠BED - ∠CEF = 90^{\circ}$。
　　$\because BE$平分$∠ABC$，而在等腰直角$△ABC$中，$∠ABC = 45^{\circ}$，$\therefore ∠C，BE = ∠ABE = 22.5^{\circ}$，$\therefore ∠ABD = ∠DBE - ∠ABE = 22.5^{\circ}$，$\therefore ∠CAF = 22.5^{\circ}$，$\therefore ∠ACE = ∠CEF - ∠CAF = 22.5^{\circ}$，$\therefore ∠ACE = ∠CAF$，$\therefore AE = CE$。
　$\because AD = 1$，$\therefore AE = CE=\sqrt{2}CF=\sqrt{2}AD=\sqrt{2}$，$\therefore BD = DE = AE + AD=\sqrt{2}+1$。在$Rt△BDE$中，$BE=\sqrt{2}BD = 2+\sqrt{2}$，$\therefore S_{△BCE}=\frac{1}{2}BE\cdot CE=\frac{1}{2}×(2+\sqrt{2})×\sqrt{2}=\sqrt{2}+1$。