《2025年轻松暑假复习加预习中国海洋大学出版社七年级数学鲁教版54制》

直角三角形
趣味数学：
盲人分袜子的方法
有两位盲人,他们各自买了两双黑袜和两双白袜,八双袜子的布质、大小完全相同,而每双袜子都由一张商标纸连着.两位盲人不小心将八双袜子混在了一起.他们两人怎样才能取回各自的黑袜和白袜呢？ 

答案： 【解析】：因为每双袜子都由一张商标纸连着，所以可以把每双袜子分开，两人各拿一只。这样对于每一双袜子，两位盲人都能拿到其中一只。由于总共是两双黑袜和两双白袜，按照这种分法，最后两人拿到的袜子组合起来就会是各自的两双黑袜和两双白袜。
【答案】：把每双袜子分开，两人各拿一只，最后两人都能取回各自的黑袜和白袜。

5. (2020·玉林)下列命题中,其逆命题是真命题的是(B)
A.对顶角相等
B.两直线平行,同位角相等
C.全等三角形的对应角相等
D.正方形的四个角都相等

答案： B

6. 如图,已知在$\triangle ABC$中,$AB= AC$,$∠C= 30^{\circ }$,$AB⊥AD$,则下列关系式正确的为(B)


A.$BD= CD$
B.$BD= 2CD$
C.$BD= 3CD$
D.$BD= 4CD$

答案： B

7. 将一副三角尺按如图所示的方式叠放在一起,若$AB= 14cm$,则阴影部分的面积是____$\frac{49}{2}$$cm^{2}$.

答案： $\frac{49}{2}$

8. (2023·吉林)如图,在$Rt\triangle ABC$中,$∠C= 90^{\circ }$,$BC\lt AC$.点D,E分别在边AB,BC上,连接DE,将$\triangle BDE$沿DE折叠,点B的对应点为点$B'$.若点$B'$刚好落在边AC上,$∠CB'E= 30^{\circ }$,$CE= 3$,则BC的长为____9.

答案： 9

9. 如图,$∠AOP= ∠BOP= 15^{\circ }$,$PC// OA$,$PD⊥OA$,若$PC= 4$,则PD的长是____.
2

答案： 2

10. 如图所示,水池中离岸边D点1.5m的C处,直立长着一根芦苇,出水部分BC的长是0.5m,把芦苇拉到岸边,它的顶端B恰好会在D处,求水深AC.
解:设$AC = x$,则$AD = AB =$$x + 0.5$,
由勾股定理得:$x^{2}+1.5^{2}=(x + 0.5)^{2}$,
解得$x =$$2$.
即水深$AC$为$2m$.

答案： 解:设$AC = x$,则$AD = AB = x + 0.5$,
由勾股定理得:$x^{2}+1.5^{2}=(x + 0.5)^{2}$,
解得$x = 2$.
即水深$AC$为$2m$.

11. $\triangle ABC$的各边长分别为$a = m^{2}-n^{2}$，$b = 2mn$，$c = m^{2}+n^{2}$。
求证：$\triangle ABC$是直角三角形。
证明：$\because (m^{2}+n^{2})^{2}=m^{4}+2m^{2}n^{2}+n^{4}$，
$(m^{2}-n^{2})^{2}+(2mn)^{2}=m^{4}+n^{4}-2m^{2}n^{2}+4m^{2}n^{2}=m^{4}+2m^{2}n^{2}+n^{4}$，
$\therefore (m^{2}-n^{2})^{2}+(2mn)^{2}=(m^{2}+n^{2})^{2}$，
即$a^{2}+b^{2}=c^{2}$。
$\therefore \triangle ABC$为直角三角形。

答案： 证明:$\because (m^{2}+n^{2})^{2}=m^{4}+2m^{2}n^{2}+n^{4}$,
$(m^{2}-n^{2})^{2}+(2mn)^{2}=m^{4}+n^{4}-2m^{2}n^{2}+4m^{2}n^{2}=m^{4}+2m^{2}n^{2}+n^{4}$,
$\therefore (m^{2}-n^{2})^{2}+(2mn)^{2}=(m^{2}+n^{2})^{2}$,
即$a^{2}+b^{2}=c^{2}$.
$\therefore \triangle ABC$为直角三角形.

12. 如图所示,在$\triangle ABC$中,$AB= CB$,$∠ABC= 90^{\circ }$,F为AB延长线上一点,点E在BC上,且$AE= CF$.
(1) 求证：$Rt\triangle ABE≌Rt\triangle CBF$；
证明:在$Rt\triangle ABE$和$Rt\triangle CBF$中,
$\left\{\begin{array}{l} AB = CB,\\ AE = CF,\end{array}\right.$
$\therefore$ $Rt\triangle ABE\cong Rt\triangle CBF(HL)$.
(2) 若$∠CAE= 30^{\circ }$,求$∠ACF$的度数.
解:$\because Rt\triangle ABE\cong Rt\triangle CBF$,$\angle CAE = 30^{\circ}$,
$\therefore$ $\angle BAE = \angle BCF = 15^{\circ}$.
$\therefore$ $\angle ACF = \angle ACB+\angle BCF = 15^{\circ}+45^{\circ}=60^{\circ}$.

答案：
(1) 证明:在$Rt\triangle ABE$和$Rt\triangle CBF$中,
$\left\{\begin{array}{l} AB = CB,\\ AE = CF,\end{array}\right.$
$\therefore Rt\triangle ABE\cong Rt\triangle CBF(HL)$.
(2) 解:$\because Rt\triangle ABE\cong Rt\triangle CBF$,$\angle CAE = 30^{\circ}$,
$\therefore \angle BAE = \angle BCF = 15^{\circ}$.
$\therefore \angle ACF = \angle ACB+\angle BCF = 15^{\circ}+45^{\circ}=60^{\circ}$.

13. 如图所示,在$\triangle ABC$中,$AB= BC$,$BE⊥AC$于点E,$AD⊥BC$于点D,$∠BAD= 45^{\circ }$,AD与BE交于点F,连接CF.
(1)求证：$BF= 2AE$；
证明：$\because AD\perp BC$,$\angle BAD = 45^{\circ}$,
$\therefore \angle ABD = \angle BAD = 45^{\circ}$.$\therefore AD = BD$.
$\because AD\perp BC$,$BE\perp AC$,
$\therefore \angle CAD+\angle ACD = 90^{\circ}$,
$\angle CBE+\angle ACD = 90^{\circ}$.
$\therefore \angle CAD = \angle CBE$.
又$\because \angle CDA = \angle FDB = 90^{\circ}$,
$\therefore \triangle ADC\cong \triangle BDF$.
$\therefore AC = BF$.
$\because AB = BC$,$BE\perp AC$,
$\therefore AE = EC$,即$AC = 2AE$.$\therefore BF = 2AE$.
(2)若$CD= \sqrt {2}$,求AD的长.
$\because \triangle ADC\cong \triangle BDF$,$\therefore DF = CD =$$\sqrt{2}$.
$\therefore$ 在$Rt\triangle CDF$中,$CF = \sqrt{DF^{2}+CD^{2}}=\sqrt{(\sqrt{2})^{2}+(\sqrt{2})^{2}} =$$2$.
$\because BE\perp AC$,$AE = EC$,$\therefore AF = FC =$$2$.
$\therefore AD = AF + DF =$$2+\sqrt{2}$.

答案： 解:
(1) 证明:$\because AD\perp BC$,$\angle BAD = 45^{\circ}$,
$\therefore \angle ABD = \angle BAD = 45^{\circ}$.$\therefore AD = BD$.
$\because AD\perp BC$,$BE\perp AC$,
$\therefore \angle CAD+\angle ACD = 90^{\circ}$,
$\angle CBE+\angle ACD = 90^{\circ}$.
$\therefore \angle CAD = \angle CBE$.
又$\because \angle CDA = \angle FDB = 90^{\circ}$,
$\therefore \triangle ADC\cong \triangle BDF$.
$\therefore AC = BF$.
$\because AB = BC$,$BE\perp AC$,
$\therefore AE = EC$,即$AC = 2AE$.$\therefore BF = 2AE$.
(2) $\because \triangle ADC\cong \triangle BDF$,$\therefore DF = CD = \sqrt{2}$.
$\therefore$ 在$Rt\triangle CDF$中,$CF = \sqrt{DF^{2}+CD^{2}}=\sqrt{(\sqrt{2})^{2}+(\sqrt{2})^{2}} = 2$.
$\because BE\perp AC$,$AE = EC$,$\therefore AF = FC = 2$.
$\therefore AD = AF + DF = 2+\sqrt{2}$.