2025年全程检测单元测试卷八年级数学上册沪教版
注:目前有些书本章节名称可能整理的还不是很完善,但都是按照顺序排列的,请同学们按照顺序仔细查找。练习册 2025年全程检测单元测试卷八年级数学上册沪教版 答案主要是用来给同学们做完题方便对答案用的,请勿直接抄袭。
1. 有些汉字是轴对称图形.下面4个汉字中,可以看作是轴对称图形的是(

A.诚
B.信
C.友
D.善
D
).A.诚
B.信
C.友
D.善
答案:
20.
(1)$\because BF$平分$\angle ABD$,$\therefore \angle ABF = \angle 1$. $\because \angle 1 = \angle 3$,$\therefore \angle ABF = \angle 3$. $\therefore AB // CD$.
(2)$\angle BED$的大小是一个定值,$\angle BED = 90^{\circ}$.理由:$\because BF$平分$\angle ABD$,$DE$平分$\angle BDC$,$\therefore \angle 1 = \frac{1}{2}\angle ABD$,$\angle 2 = \frac{1}{2}\angle BDF$. $\because AB // CD$,$\therefore \angle ABD + \angle BDF = 180^{\circ}$. $\therefore \angle 1 + \angle 2 = \frac{1}{2}(\angle ABD + \angle BDF) = \frac{1}{2} × 180^{\circ} = 90^{\circ}$. $\therefore \angle BED = 180^{\circ} - (\angle 1 + \angle 2) = 180^{\circ} - 90^{\circ} = 90^{\circ}$.
(1)$\because BF$平分$\angle ABD$,$\therefore \angle ABF = \angle 1$. $\because \angle 1 = \angle 3$,$\therefore \angle ABF = \angle 3$. $\therefore AB // CD$.
(2)$\angle BED$的大小是一个定值,$\angle BED = 90^{\circ}$.理由:$\because BF$平分$\angle ABD$,$DE$平分$\angle BDC$,$\therefore \angle 1 = \frac{1}{2}\angle ABD$,$\angle 2 = \frac{1}{2}\angle BDF$. $\because AB // CD$,$\therefore \angle ABD + \angle BDF = 180^{\circ}$. $\therefore \angle 1 + \angle 2 = \frac{1}{2}(\angle ABD + \angle BDF) = \frac{1}{2} × 180^{\circ} = 90^{\circ}$. $\therefore \angle BED = 180^{\circ} - (\angle 1 + \angle 2) = 180^{\circ} - 90^{\circ} = 90^{\circ}$.
2. 如图1,△ABC和△DEF关于直线l对称,点B的对称点是(

A.点C
B.点D
C.点E
D.点F
C
).A.点C
B.点D
C.点E
D.点F
答案:
21.
(1)联立方程$\begin{cases}y = 2x + 1, \\y = -x - 2.\end{cases}$解得$\begin{cases}x = -1, \\y = -1.\end{cases}$ $\therefore$点$P$的坐标为$(-1,-1)$.
(2)把$y = 0$代入$y_1 = 2x + 1$,得$2x + 1 = 0$.解得$x = -\frac{1}{2}$. $\therefore$点$C$的坐标为$(-\frac{1}{2},0)$.把$y = 0$代入$y_2 = -x - 2$,得$-x - 2 = 0$.解得$x = -2$. $\therefore$点$D$的坐标为$(-2,0)$. $\therefore S_{\triangle PCD} = \frac{1}{2}CD \cdot |y_P| = \frac{1}{2} × \left[-\frac{1}{2} - (-2)\right] × 1 = \frac{3}{4}$.
(3)由图象可知,在直线$x = -1$的右侧,直线$l_1$在直线$l_2$上方,$\therefore$当$x \geq -1$时,$y_1 \geq y_2$.
(1)联立方程$\begin{cases}y = 2x + 1, \\y = -x - 2.\end{cases}$解得$\begin{cases}x = -1, \\y = -1.\end{cases}$ $\therefore$点$P$的坐标为$(-1,-1)$.
(2)把$y = 0$代入$y_1 = 2x + 1$,得$2x + 1 = 0$.解得$x = -\frac{1}{2}$. $\therefore$点$C$的坐标为$(-\frac{1}{2},0)$.把$y = 0$代入$y_2 = -x - 2$,得$-x - 2 = 0$.解得$x = -2$. $\therefore$点$D$的坐标为$(-2,0)$. $\therefore S_{\triangle PCD} = \frac{1}{2}CD \cdot |y_P| = \frac{1}{2} × \left[-\frac{1}{2} - (-2)\right] × 1 = \frac{3}{4}$.
(3)由图象可知,在直线$x = -1$的右侧,直线$l_1$在直线$l_2$上方,$\therefore$当$x \geq -1$时,$y_1 \geq y_2$.
3. 如图2,若△ABC与△A'B'C'关于直线MN对称,对应点所连线段BB'与直线MN交于点O.下列说法不一定正确的是().

A.AC=A'C'
B.AB//B'C'
C.CC'⊥MN
D.AO=A'O
A.AC=A'C'
B.AB//B'C'
C.CC'⊥MN
D.AO=A'O
答案:
22.
(1)根据题意,两个函数均为一次函数,设$y = kt + b$,将$(10,10)$,$(20,20)$代入,得$\begin{cases}10k + b = 10, \\20k + b = 20.\end{cases}$解得$\begin{cases}k = 1, \\b = 0.\end{cases}$ $\therefore y = t$.设$e = ms + n$,将$(0,100)$,$(100,70)$代入,得$\begin{cases}n = 100, \\100m + n = 70.\end{cases}$解得$\begin{cases}m = -0.3, \\n = 100.\end{cases}$ $\therefore e = -0.3s + 100$.
(2)由题意得,电动汽车首先在充满电的情况下行驶了140 km,$\therefore$当$s = 140$时,$e = -0.3s + 100 = -0.3 × 140 + 100 = 58$. $\therefore$到服务区充电时,电量显示为58%.假设充电充了$t$ min,应增加电量$t\%$,则再次出发时电量为$(58 + t)\%$,走完剩余路程$360 - 140 = 220$ (km)应耗电量为$0.3\% × 220 = 66\%$,而剩余电量为30%,据此可得$30 = 58 + t - 66$,解得$t = 38$.答:电动汽车在服务区充电38 min.
(1)根据题意,两个函数均为一次函数,设$y = kt + b$,将$(10,10)$,$(20,20)$代入,得$\begin{cases}10k + b = 10, \\20k + b = 20.\end{cases}$解得$\begin{cases}k = 1, \\b = 0.\end{cases}$ $\therefore y = t$.设$e = ms + n$,将$(0,100)$,$(100,70)$代入,得$\begin{cases}n = 100, \\100m + n = 70.\end{cases}$解得$\begin{cases}m = -0.3, \\n = 100.\end{cases}$ $\therefore e = -0.3s + 100$.
(2)由题意得,电动汽车首先在充满电的情况下行驶了140 km,$\therefore$当$s = 140$时,$e = -0.3s + 100 = -0.3 × 140 + 100 = 58$. $\therefore$到服务区充电时,电量显示为58%.假设充电充了$t$ min,应增加电量$t\%$,则再次出发时电量为$(58 + t)\%$,走完剩余路程$360 - 140 = 220$ (km)应耗电量为$0.3\% × 220 = 66\%$,而剩余电量为30%,据此可得$30 = 58 + t - 66$,解得$t = 38$.答:电动汽车在服务区充电38 min.
4. 如图3,若整个图案的面积为28π,则阴影部分的面积为().

A.28π
B.14π
C.7π
D.3.5π
A.28π
B.14π
C.7π
D.3.5π
答案:
B
5. 小王准备在街道旁建一个送奶站,向居民区A,B提供牛奶,要使A,B两小区到送奶站的距离的和最小,则下列所作送奶站C的位置正确的是().

答案:
C
6. 在联欢会上,甲、乙、丙三名选手在玩抢凳子游戏,规则是:如图4,三人分别站在△ABC三个顶点的位置上,在三角形内部放一张凳子,音乐停止后立即开始抢凳子,谁先抢到凳子谁获胜.为使游戏公平,凳子应放在△ABC的().

A.三条中线的交点
B.三边垂直平分线的交点
C.三条角平分线的交点
D.三条高的交点
A.三条中线的交点
B.三边垂直平分线的交点
C.三条角平分线的交点
D.三条高的交点
答案:
B
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