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15. 如图,在$Rt△ ABC$中,$∠ACB=90^{\circ },AC=2,BC=\sqrt{2}$.分别
以 AB,AC,BC 为边向外作正方形 ABDE、正方形 ACFG、
正方形 BCMN,连接 GE,DN,则图中阴影部分的总面积是
.
以 AB,AC,BC 为边向外作正方形 ABDE、正方形 ACFG、
正方形 BCMN,连接 GE,DN,则图中阴影部分的总面积是
.
答案
$2\sqrt{2}$
解析
1. 计算$Rt△ ABC$的面积：$S_{△ ABC}=\frac{1}{2}AC· BC=\frac{1}{2}×2×\sqrt{2}=\sqrt{2}$；
2. 证明$S_{△ AGE}=S_{△ ABC}$：过点$E$作$EH⊥ GA$的延长线于$H$，由$∠ EAH+∠ CAB=90°$，$∠ CAB+∠ ABC=90°$，得$∠ EAH=∠ ABC$，结合$AE=AB$，$∠ EHA=∠ ACB=90°$，可证$△ EHA≌△ BCA(AAS)$，则$EH=BC=\sqrt{2}$，故$S_{△ AGE}=\frac{1}{2}AG· EH=\frac{1}{2}×2×\sqrt{2}=\sqrt{2}$；
3. 同理可证$S_{△ BDN}=S_{△ ABC}=\sqrt{2}$；
4. 阴影部分总面积为$\sqrt{2}+\sqrt{2}=2\sqrt{2}$。
2. 证明$S_{△ AGE}=S_{△ ABC}$：过点$E$作$EH⊥ GA$的延长线于$H$，由$∠ EAH+∠ CAB=90°$，$∠ CAB+∠ ABC=90°$，得$∠ EAH=∠ ABC$，结合$AE=AB$，$∠ EHA=∠ ACB=90°$，可证$△ EHA≌△ BCA(AAS)$，则$EH=BC=\sqrt{2}$，故$S_{△ AGE}=\frac{1}{2}AG· EH=\frac{1}{2}×2×\sqrt{2}=\sqrt{2}$；
3. 同理可证$S_{△ BDN}=S_{△ ABC}=\sqrt{2}$；
4. 阴影部分总面积为$\sqrt{2}+\sqrt{2}=2\sqrt{2}$。
三、解答题(共75分)
16.(6分)计算:
(1)$\sqrt{27}÷\frac{\sqrt{3}}{2}×2\sqrt{2}-6\sqrt{2}$;
(2)$\sqrt{3}×(\sqrt{2}-\sqrt{3})-\sqrt{24}-|\sqrt{6}-3|$.
16.(6分)计算:
(1)$\sqrt{27}÷\frac{\sqrt{3}}{2}×2\sqrt{2}-6\sqrt{2}$;
(2)$\sqrt{3}×(\sqrt{2}-\sqrt{3})-\sqrt{24}-|\sqrt{6}-3|$.
答案
解：
(1) $\sqrt{27}÷\frac{\sqrt{3}}{2}×2\sqrt{2}-6\sqrt{2}$
$=3\sqrt{3}×\frac{2}{\sqrt{3}}×2\sqrt{2}-6\sqrt{2}$
$=12\sqrt{2}-6\sqrt{2}$
$=6\sqrt{2}$
(2) $\sqrt{3}×(\sqrt{2}-\sqrt{3})-\sqrt{24}-|\sqrt{6}-3|$
$=\sqrt{6}-3-2\sqrt{6}-(3-\sqrt{6})$
$=\sqrt{6}-3-2\sqrt{6}-3+\sqrt{6}$
$=-6$
(1) $\sqrt{27}÷\frac{\sqrt{3}}{2}×2\sqrt{2}-6\sqrt{2}$
$=3\sqrt{3}×\frac{2}{\sqrt{3}}×2\sqrt{2}-6\sqrt{2}$
$=12\sqrt{2}-6\sqrt{2}$
$=6\sqrt{2}$
(2) $\sqrt{3}×(\sqrt{2}-\sqrt{3})-\sqrt{24}-|\sqrt{6}-3|$
$=\sqrt{6}-3-2\sqrt{6}-(3-\sqrt{6})$
$=\sqrt{6}-3-2\sqrt{6}-3+\sqrt{6}$
$=-6$
17.(8分)先观察解题过程,再解决以下问题:
比较$\sqrt{3}-\sqrt{2}$与$\sqrt{2}-1$的大小.
解:$\because (\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})=1,(\sqrt{2}-1)(\sqrt{2}+1)=1,$
$\therefore \sqrt{3}-\sqrt{2}=\frac{1}{\sqrt{3}+\sqrt{2}},\sqrt{2}-1=\frac{1}{\sqrt{2}+1}.$
又$\because \sqrt{3}+\sqrt{2}＞\sqrt{2}+1,\therefore \sqrt{3}-\sqrt{2}＜\sqrt{2}-1.$
(1)比较$2-\sqrt{3}$与$\sqrt{3}-\sqrt{2}$的大小;
(2)直接比较$\sqrt{n+1}-\sqrt{n}$与$\sqrt{n}-\sqrt{n-1}$的大小.
比较$\sqrt{3}-\sqrt{2}$与$\sqrt{2}-1$的大小.
解:$\because (\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})=1,(\sqrt{2}-1)(\sqrt{2}+1)=1,$
$\therefore \sqrt{3}-\sqrt{2}=\frac{1}{\sqrt{3}+\sqrt{2}},\sqrt{2}-1=\frac{1}{\sqrt{2}+1}.$
又$\because \sqrt{3}+\sqrt{2}＞\sqrt{2}+1,\therefore \sqrt{3}-\sqrt{2}＜\sqrt{2}-1.$
(1)比较$2-\sqrt{3}$与$\sqrt{3}-\sqrt{2}$的大小;
(2)直接比较$\sqrt{n+1}-\sqrt{n}$与$\sqrt{n}-\sqrt{n-1}$的大小.
答案
解：
(1) $\because (2-\sqrt{3})(2+\sqrt{3})=4-3=1,(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})=3-2=1,$
$\therefore 2-\sqrt{3}=\frac{1}{2+\sqrt{3}},\sqrt{3}-\sqrt{2}=\frac{1}{\sqrt{3}+\sqrt{2}}.$
又$\because 2+\sqrt{3}＞\sqrt{3}+\sqrt{2},\therefore 2-\sqrt{3}＜\sqrt{3}-\sqrt{2}.$
(2) $\because (\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})=(n+1)-n=1,(\sqrt{n}-\sqrt{n-1})(\sqrt{n}+\sqrt{n-1})=n-(n-1)=1,$
$\therefore \sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}},\sqrt{n}-\sqrt{n-1}=\frac{1}{\sqrt{n}+\sqrt{n-1}}.$
又$\because \sqrt{n+1}+\sqrt{n}＞\sqrt{n}+\sqrt{n-1}(n≥1且n为整数),$
$\therefore \sqrt{n+1}-\sqrt{n}＜\sqrt{n}-\sqrt{n-1}.$
(1) $\because (2-\sqrt{3})(2+\sqrt{3})=4-3=1,(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})=3-2=1,$
$\therefore 2-\sqrt{3}=\frac{1}{2+\sqrt{3}},\sqrt{3}-\sqrt{2}=\frac{1}{\sqrt{3}+\sqrt{2}}.$
又$\because 2+\sqrt{3}＞\sqrt{3}+\sqrt{2},\therefore 2-\sqrt{3}＜\sqrt{3}-\sqrt{2}.$
(2) $\because (\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})=(n+1)-n=1,(\sqrt{n}-\sqrt{n-1})(\sqrt{n}+\sqrt{n-1})=n-(n-1)=1,$
$\therefore \sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}},\sqrt{n}-\sqrt{n-1}=\frac{1}{\sqrt{n}+\sqrt{n-1}}.$
又$\because \sqrt{n+1}+\sqrt{n}＞\sqrt{n}+\sqrt{n-1}(n≥1且n为整数),$
$\therefore \sqrt{n+1}-\sqrt{n}＜\sqrt{n}-\sqrt{n-1}.$
