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19. 如图,在平面直角坐标系中,点$A_{1},A_{2},A_{3},...都在x$轴上,点$B_{1},B_{2},B_{3},...都在直线y= x$上,$△OA_{1}B_{1},△B_{1}A_{1}A_{2},△B_{2}A_{2}A_{3},△B_{3}A_{3}A_{4}$都是等腰直角三角形.若$OA_{1}= 1$,则点$B_{218}$的坐标是
$(2^{217},2^{217})$
.答案
$(2^{217},2^{217})$
20. 计算:
(1)$(\sqrt {2}-3)^{2}-\sqrt {2}$;
(2)$\sqrt {3}×\sqrt {2}-\sqrt {12}÷\sqrt {8}$.
(1)$(\sqrt {2}-3)^{2}-\sqrt {2}$;
(2)$\sqrt {3}×\sqrt {2}-\sqrt {12}÷\sqrt {8}$.
答案
解:(1)原式$=2 - 6\sqrt{2} + 9 - \sqrt{2} = 11 - 7\sqrt{2}$.
(2)原式$=\sqrt{6} - \frac{3}{2}\sqrt{6} - \frac{\sqrt{6}}{2} = \frac{\sqrt{6}}{2}$.
(2)原式$=\sqrt{6} - \frac{3}{2}\sqrt{6} - \frac{\sqrt{6}}{2} = \frac{\sqrt{6}}{2}$.
21. 已知$y与x-1$成正比例,且当$x= -1$时,$y= 4$.
(1)求$y关于x$的函数解析式;
(2)点$M(x_{1},y_{1}),N(x_{2},y_{2})$在(1)中函数的图象上,若$x_{1}＞x_{2}$,则$y_{1}$
(3)将(1)中函数的图象向下平移 4 个单位长度,得到的新图象与$x$轴、$y轴分别交于点A,B$,求$△AOB$的面积.
(1)求$y关于x$的函数解析式;
(2)点$M(x_{1},y_{1}),N(x_{2},y_{2})$在(1)中函数的图象上,若$x_{1}＞x_{2}$,则$y_{1}$
＜
$y_{2}$(填“＞”“＜”或“=”);(3)将(1)中函数的图象向下平移 4 个单位长度,得到的新图象与$x$轴、$y轴分别交于点A,B$,求$△AOB$的面积.
答案
解:(1)设$y$关于$x$的函数解析式为$y = k(x - 1)$.
$\because$当$x = -1$时,$y = 4$,$\therefore 4 = k×(-1 - 1)$. 解得$k = -2$.$\therefore y$关于$x$的函数解析式为$y = -2(x - 1) = -2x + 2$. (2)$＜$ (3)将$y = -2x + 2$的图象向下平移 4 个单位长度,所得图象的解析式为$y = -2x - 2$. 当$x = 0$时,$y = -2$;当$y = 0$时,$x = -1$,$\therefore A(-1,0)$,$B(0,-2)$.$\therefore S_{\triangle AOB} = \frac{1}{2}×1×2 = 1$.
$\because$当$x = -1$时,$y = 4$,$\therefore 4 = k×(-1 - 1)$. 解得$k = -2$.$\therefore y$关于$x$的函数解析式为$y = -2(x - 1) = -2x + 2$. (2)$＜$ (3)将$y = -2x + 2$的图象向下平移 4 个单位长度,所得图象的解析式为$y = -2x - 2$. 当$x = 0$时,$y = -2$;当$y = 0$时,$x = -1$,$\therefore A(-1,0)$,$B(0,-2)$.$\therefore S_{\triangle AOB} = \frac{1}{2}×1×2 = 1$.
22. 如图,已知在$△ABC$中,$BC= 13,D是线段AC$上的一点,连接$BD,CD= 5,BD= 12$.
(1)求证:$BD⊥AC$;
证明:$\because CD^{2} + BD^{2} = 5^{2} + 12^{2} = 169$,$BC^{2} = 13^{2} = 169$,$\therefore CD^{2} + BD^{2} = BC^{2}$.$\therefore \triangle BDC$为直角三角形.$\therefore$
(2)若$S_{△ABC}= 48$,求$△ABC$的周长.
解:$\because S_{\triangle ABC} = 48$,$BD = 12$,$BD⊥AC$,$\therefore AC =$
(1)求证:$BD⊥AC$;
证明:$\because CD^{2} + BD^{2} = 5^{2} + 12^{2} = 169$,$BC^{2} = 13^{2} = 169$,$\therefore CD^{2} + BD^{2} = BC^{2}$.$\therefore \triangle BDC$为直角三角形.$\therefore$
$BD⊥AC$
.(2)若$S_{△ABC}= 48$,求$△ABC$的周长.
解:$\because S_{\triangle ABC} = 48$,$BD = 12$,$BD⊥AC$,$\therefore AC =$
8
.$\therefore AD = AC - CD = 8 - 5 =$3
.$\because BD⊥AC$,$\therefore AB = \sqrt{BD^{2} + AD^{2}} = \sqrt{12^{2} + 3^{2}} =$$3\sqrt{17}$
.$\therefore C_{\triangle ABC} = AB + AC + BC = 3\sqrt{17} + 8 + 13 =$$3\sqrt{17} + 21$
.答案
(1)证明:$\because CD^{2} + BD^{2} = 5^{2} + 12^{2} = 169$,$BC^{2} = 13^{2} = 169$,$\therefore CD^{2} + BD^{2} = BC^{2}$.$\therefore \triangle BDC$为直角三角形.$\therefore BD⊥AC$. (2)解:$\because S_{\triangle ABC} = 48$,$BD = 12$,$BD⊥AC$,$\therefore AC = 8$.$\therefore AD = AC - CD = 8 - 5 = 3$.$\because BD⊥AC$,$\therefore AB = \sqrt{BD^{2} + AD^{2}} = \sqrt{12^{2} + 3^{2}} = 3\sqrt{17}$.$\therefore C_{\triangle ABC} = AB + AC + BC = 3\sqrt{17} + 8 + 13 = 3\sqrt{17} + 21$.
23. 如图,在$□ ABCD$中,$E,F是对角线AC$上的两点,且$AE= CF$,连接$DE,DF,BE,BF$.
(1)求证:四边形$BEDF$是平行四边形.
(2)若$AD⊥DF,DF= 5,AC= 14,∠DAC= 30^{\circ }$.
①求线段$EF$的长;
②求四边形$BEDF$的面积.
(1)求证:四边形$BEDF$是平行四边形.
(2)若$AD⊥DF,DF= 5,AC= 14,∠DAC= 30^{\circ }$.
①求线段$EF$的长;
②求四边形$BEDF$的面积.
答案
(1)证明:$\because$四边形$ABCD$是平行四边形,$\therefore AD = BC$,$AD// BC$.$\therefore ∠DAE = ∠BCF$. 在$\triangle DAE$和$\triangle BCF$中,$\begin{cases}AD = BC,\\∠DAE = ∠BCF,\\AE = CF,\end{cases}$ $\therefore \triangle DAE ≌ \triangle BCF(SAS)$.$\therefore DE = BF$,$∠AED = ∠CFB$.$\because$点$E$,$F$经过直线$AC$,$\therefore ∠DEF = ∠BFE$.$\therefore DE// BF$.$\therefore$四边形$BEDF$是平行四边形.
(2)解:①$\because AD⊥DF$,$∠DAC = 30^{\circ}$,$DF = 5$,$\therefore AF = 2DF = 10$.$\therefore CF = AC - AF = 14 - 10 = 4$.$\therefore AE = CF = 4$.$\therefore EF = AF - AE = 10 - 4 = 6$.
②如答图,过点$D$作$DM⊥EF$交$EF$于点$M$.
$\because AD⊥DF$,$AF = 10$,$DF = 5$,$\therefore AD = \sqrt{10^{2} - 5^{2}} = 5\sqrt{3}$.$\because ∠DAC = 30^{\circ}$,$\therefore DM = \frac{1}{2}AD = \frac{5\sqrt{3}}{2}$.$\therefore S_{四边形BEDF} = 2S_{\triangle DEF} = 2×\frac{1}{2}×6×\frac{5\sqrt{3}}{2} = 15\sqrt{3}$.
