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19. (6分)(1)如图1,直线EF分别与直线AB,CD交于点E,F,EM平分∠BEF,FN平分∠CFE,且EM//FN,求证:AB//CD;
(2)如图2,已知AB//CD,∠1 = ∠2,求证:∠BPM = ∠CMP.
(2)如图2,已知AB//CD,∠1 = ∠2,求证:∠BPM = ∠CMP.
答案
19. 【点拨】本题考查平行线的判定和性质,角平分线的定义.
【解析】(1)证明:$\because EM // FN$,$\therefore ∠ FEM = ∠ EFN$.
$\because EM$平分$∠ BEF$,$FN$平分$∠ CFE$,
$\therefore ∠ BEF = 2∠ FEM$,$∠ CFE = 2∠ EFN$,
$\therefore ∠ BEF = ∠ CFE$,$\therefore AB // CD$.
(2)证明:如题图2,延长BP,DC交于点Q.
$\because AB // CD$,$\therefore ∠ 1 = ∠ Q$.
$\because ∠ 1 = ∠ 2$,$\therefore ∠ 2 = ∠ Q$,$\therefore PQ // CM$,
$\therefore ∠ BPM = ∠ CMP$.
【解析】(1)证明:$\because EM // FN$,$\therefore ∠ FEM = ∠ EFN$.
$\because EM$平分$∠ BEF$,$FN$平分$∠ CFE$,
$\therefore ∠ BEF = 2∠ FEM$,$∠ CFE = 2∠ EFN$,
$\therefore ∠ BEF = ∠ CFE$,$\therefore AB // CD$.
(2)证明:如题图2,延长BP,DC交于点Q.
$\because AB // CD$,$\therefore ∠ 1 = ∠ Q$.
$\because ∠ 1 = ∠ 2$,$\therefore ∠ 2 = ∠ Q$,$\therefore PQ // CM$,
$\therefore ∠ BPM = ∠ CMP$.
20. (6分)(1)如图1,分别把两个边长为1 cm 的小正方形沿一条对角线裁成4个小三角形拼成一个大正方形,则大正方形的边长为
(2)如图2,3是4×4的网格,在坐标平面内,已知$A(-1,0)$,结合上面的知识完成下列问题:
①在图2中建立平面直角坐标系;(坐标轴在网格线所在的直线上,不写作法)
②在现有网格中将点A先向右平移2个单位长度,再向上平移2个单位长度得到点B,则点B的坐标为
③请在图3中画出一个面积为8的正方形.
$\sqrt{2}$
cm;(2)如图2,3是4×4的网格,在坐标平面内,已知$A(-1,0)$,结合上面的知识完成下列问题:
①在图2中建立平面直角坐标系;(坐标轴在网格线所在的直线上,不写作法)
②在现有网格中将点A先向右平移2个单位长度,再向上平移2个单位长度得到点B,则点B的坐标为
$(1,2)$
,$AB=$$2\sqrt{2}$
;③请在图3中画出一个面积为8的正方形.
答案
20. 【点拨】本题考查二次根式的应用和平面直角坐标系.
【解析】(1)由题意知,大正方形的面积为$2\ \mathrm{cm}^2$,则大正方形的边长为$\sqrt{2}\ \mathrm{cm}$.故答案为$\sqrt{2}$.
(2)①如图
所示为建立的平面直角坐标系.②如图
,点B的坐标为$(1,2)$,$AB=2\sqrt{2}$.故答案为$(1,2)$,$2\sqrt{2}$.③如图
所示,正方形DEFG即为所求.21.（6分）【阅读理解】请阅读下面材料，并完成相应的任务.
【问题提出】$\sqrt{2}$有多大呢？
【问题解决】$\because 1^2 = 1,2^2 = 4,\therefore 1 ＜ \sqrt{2} ＜ 2$.
$\because 1.4^2 = 1.96,1.5^2 = 2.25,\therefore 1.4 ＜ \sqrt{2} ＜ 1.5$.
$\because 1.41^2 = 1.988\ 1,1.42^2 = 2.016\ 4,\therefore 1.41 ＜ \sqrt{2} ＜ 1.42$.
又$\because 1.414^2 = 1.999\ 396,1.415^2 = 2.002\ 225,\therefore 1.414 ＜ \sqrt{2} ＜ 1.415$.
(1)$\sqrt{2}$的近似值为________；(结果保留两位小数)
(2)用上述方法估算$\sqrt{11}$的计算值；(结果保留两位小数)
(3)若$\sqrt{17} + \frac{4}{5}$的整数部分为$a$，小数部分为$b$，$x,y$均为有理数，且满足$a + 5b = \sqrt{17}x + y$，求$xy$的值.
【问题提出】$\sqrt{2}$有多大呢？
【问题解决】$\because 1^2 = 1,2^2 = 4,\therefore 1 ＜ \sqrt{2} ＜ 2$.
$\because 1.4^2 = 1.96,1.5^2 = 2.25,\therefore 1.4 ＜ \sqrt{2} ＜ 1.5$.
$\because 1.41^2 = 1.988\ 1,1.42^2 = 2.016\ 4,\therefore 1.41 ＜ \sqrt{2} ＜ 1.42$.
又$\because 1.414^2 = 1.999\ 396,1.415^2 = 2.002\ 225,\therefore 1.414 ＜ \sqrt{2} ＜ 1.415$.
(1)$\sqrt{2}$的近似值为________；(结果保留两位小数)
(2)用上述方法估算$\sqrt{11}$的计算值；(结果保留两位小数)
(3)若$\sqrt{17} + \frac{4}{5}$的整数部分为$a$，小数部分为$b$，$x,y$均为有理数，且满足$a + 5b = \sqrt{17}x + y$，求$xy$的值.
答案
21. 【点拨】本题考查无理数的估算.
【解析】(1)$\because 1.414 ＜ \sqrt{2} ＜ 1.415$.$\therefore \sqrt{2}$的近似值为1.41.故答案为1.41.
(2)$\because 3^2 = 9$,$4^2 = 16$,$\therefore 3 ＜ \sqrt{11} ＜ 4$.
$\because 3.3^2 = 10.89$,$3.4^2 = 11.56$,$\therefore 3.3 ＜ \sqrt{11} ＜ 3.4$.
$\because 3.31^2 = 10.9561$,$3.32^2 = 11.0224$,$\therefore 3.31 ＜ \sqrt{11} ＜ 3.32$.
$\because 3.316^2 = 10.995856$,$3.317^2 = 11.002489$,$\therefore 3.316 ＜ \sqrt{11} ＜ 3.317$,$\therefore \sqrt{11} \approx 3.32$.
(3)$\because 4^2 = 16$,$5^2 = 25$,$\therefore 4 ＜ \sqrt{17} ＜ 5$.
$\because 4.1^2 = 16.81$,$4.2^2 = 17.64$,$\therefore 4.1 ＜ \sqrt{17} ＜ 4.2$.
$\because 4.11^2 = 16.8921$,$4.15^2 = 17.2225$,$\therefore 4.11 ＜ \sqrt{17} ＜ 4.15$,$\therefore \sqrt{17} + 0.8 \approx 4.9$,$\therefore a=4$,$b=\sqrt{17} + 0.8 - 4$,
$\therefore a + 5b = 4 + 5 × (\sqrt{17} + 0.8 - 4) = 5\sqrt{17} - 12 = \sqrt{17}x + y$,
$\therefore x=5$,$y=-12$,$\therefore xy=-60$.
【解析】(1)$\because 1.414 ＜ \sqrt{2} ＜ 1.415$.$\therefore \sqrt{2}$的近似值为1.41.故答案为1.41.
(2)$\because 3^2 = 9$,$4^2 = 16$,$\therefore 3 ＜ \sqrt{11} ＜ 4$.
$\because 3.3^2 = 10.89$,$3.4^2 = 11.56$,$\therefore 3.3 ＜ \sqrt{11} ＜ 3.4$.
$\because 3.31^2 = 10.9561$,$3.32^2 = 11.0224$,$\therefore 3.31 ＜ \sqrt{11} ＜ 3.32$.
$\because 3.316^2 = 10.995856$,$3.317^2 = 11.002489$,$\therefore 3.316 ＜ \sqrt{11} ＜ 3.317$,$\therefore \sqrt{11} \approx 3.32$.
(3)$\because 4^2 = 16$,$5^2 = 25$,$\therefore 4 ＜ \sqrt{17} ＜ 5$.
$\because 4.1^2 = 16.81$,$4.2^2 = 17.64$,$\therefore 4.1 ＜ \sqrt{17} ＜ 4.2$.
$\because 4.11^2 = 16.8921$,$4.15^2 = 17.2225$,$\therefore 4.11 ＜ \sqrt{17} ＜ 4.15$,$\therefore \sqrt{17} + 0.8 \approx 4.9$,$\therefore a=4$,$b=\sqrt{17} + 0.8 - 4$,
$\therefore a + 5b = 4 + 5 × (\sqrt{17} + 0.8 - 4) = 5\sqrt{17} - 12 = \sqrt{17}x + y$,
$\therefore x=5$,$y=-12$,$\therefore xy=-60$.
