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21. 计算:
(1)$3 ÷ \sqrt { 3 } × \dfrac { 1 } { \sqrt { 3 } }$;
(2)$( \sqrt { 24 } + \sqrt { 0.5 } ) - \left( \sqrt { \dfrac { 1 } { 8 } } + \sqrt { 6 } \right)$.
(1)$3 ÷ \sqrt { 3 } × \dfrac { 1 } { \sqrt { 3 } }$;
(2)$( \sqrt { 24 } + \sqrt { 0.5 } ) - \left( \sqrt { \dfrac { 1 } { 8 } } + \sqrt { 6 } \right)$.
答案
解:(1)原式$= 3 × \frac{\sqrt{3}}{3} × \frac{\sqrt{3}}{3} = \sqrt{3} × \frac{\sqrt{3}}{3} = 1$. (2)原式$= (2\sqrt{6} + \frac{\sqrt{2}}{2}) - (\frac{\sqrt{2}}{4} + \sqrt{6}) = 2\sqrt{6} - \sqrt{6} + \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{4} = \sqrt{6} + \frac{\sqrt{2}}{4}$.
22. 如图,已知等边三角形$A B C的边长是4 \sqrt { 2 }$,求它的面积.
答案
解:如答图，过点A作$AE \perp BC$于点E. 在等边三角形ABC中，$AB = AC$，$\therefore BE = CE$. $\because$ 等边三角形ABC的边长是$4\sqrt{2}$，$\therefore BE = 2\sqrt{2}$. 在$Rt\triangle ABE$中，根据勾股定理，得$AE = \sqrt{AB^{2} - BE^{2}} = \sqrt{(4\sqrt{2})^{2} - (2\sqrt{2})^{2}} = 2\sqrt{6}$，$\therefore \triangle ABC$的面积为$\frac{1}{2}BC \cdot AE = \frac{1}{2} × 4\sqrt{2} × 2\sqrt{6} = 8\sqrt{3}$.
23. 如图,四边形$A B C D$是菱形,$B E \perp A D于点E$,$B F \perp C D于点F$.求证:$D E = D F$.
证明:$\because$ 四边形ABCD是菱形，$\therefore AB = BC = CD = AD$，$\angle A = \angle C$. $\because BE \perp AD$，$BF \perp CD$，$\therefore \angle BEA = \angle BFC = 90^{\circ}$. 在$\triangle ABE$和$\triangle CBF$中，$\begin{cases} \angle BEA = \angle BFC, \\ \angle A = \angle C, \\ AB = BC, \end{cases}$ $\therefore \triangle ABE \cong \triangle CBF$(
证明:$\because$ 四边形ABCD是菱形，$\therefore AB = BC = CD = AD$，$\angle A = \angle C$. $\because BE \perp AD$，$BF \perp CD$，$\therefore \angle BEA = \angle BFC = 90^{\circ}$. 在$\triangle ABE$和$\triangle CBF$中，$\begin{cases} \angle BEA = \angle BFC, \\ \angle A = \angle C, \\ AB = BC, \end{cases}$ $\therefore \triangle ABE \cong \triangle CBF$(
AAS
). $\therefore AE = CF$. $\because AD = CD$，$\therefore AD - AE = CD - CF$，即$DE = DF$.答案
证明:$\because$ 四边形ABCD是菱形，$\therefore AB = BC = CD = AD$，$\angle A = \angle C$. $\because BE \perp AD$，$BF \perp CD$，$\therefore \angle BEA = \angle BFC = 90^{\circ}$. 在$\triangle ABE$和$\triangle CBF$中，$\begin{cases} \angle BEA = \angle BFC, \\ \angle A = \angle C, \\ AB = BC, \end{cases}$ $\therefore \triangle ABE \cong \triangle CBF(AAS)$. $\therefore AE = CF$. $\because AD = CD$，$\therefore AD - AE = CD - CF$，即$DE = DF$.
24. 如图,直线$y = a x + 6与直线y = 2 x相交于点A ( m , 4 )$,且与$x轴交于点B$,与$y轴交于点C$.
(1)求$a$和$m$的值;$a=$
(2)求$\triangle A O C$的面积.
(1)求$a$和$m$的值;$a=$
-1
，$m=$2
(2)求$\triangle A O C$的面积.
6
答案
解:(1)$\because$ 直线$y = 2x$过点$A(m, 4)$，$\therefore 4 = 2m$. 解得$m = 2$. $\therefore$ 点A的坐标为$(2, 4)$. $\because$ 直线$y = ax + 6$过点A，$\therefore 2a + 6 = 4$. 解得$a = -1$. (2)把$x = 0$代入$y = -x + 6$，得$y = 6$. $\therefore C(0, 6)$. $\therefore OC = 6$. $\therefore S_{\triangle AOC} = \frac{1}{2}OC × x_{A} = \frac{1}{2} × 6 × 2 = 6$.
25. 如图所示的条形统计图描述了某车间工人日加工零件数的情况.
(1)直接写出该车间工人日加工零件数的众数、中位数.众数为
(2)该车间工人平均每人日加工零件数大约是多少个?(结果取整数)
(3)为调动工人积极性,车间欲制订每人日加工零件数指标,多于指标者予以奖励,为了使多数工人获得奖励,应根据____制订这个指标.(选填“平均数”“众数”或“中位数”)
(1)直接写出该车间工人日加工零件数的众数、中位数.众数为
6
，中位数为6
.(2)该车间工人平均每人日加工零件数大约是多少个?(结果取整数)
6
(3)为调动工人积极性,车间欲制订每人日加工零件数指标,多于指标者予以奖励,为了使多数工人获得奖励,应根据____制订这个指标.(选填“平均数”“众数”或“中位数”)
中位数
答案
解:(1)由题意可知，该车间工人日加工零件数的众数为6，该车间一共有$4 + 5 + 8 + 9 + 6 + 4 = 36$(人)，该车间工人日加工零件数从小到大排列排在中间的两个数均为6个，故中位数为$\frac{6 + 6}{2} = 6$. (2)$\frac{1}{36} × (4 × 3 + 5 × 4 + 8 × 5 + 9 × 6 + 6 × 7 + 4 × 8) \approx 6$(个). 答：该车间工人平均每人日加工零件数大约是6个. (3)中位数
