《2025年亮点给力大试卷八年级数学上册苏科版》

25. (6分)如图①,在四边形ABCD中,$∠B= 90^{\circ },AD// BC,AB= 4,AD= 6$,动点P从点B出发,以每秒2个单位长度的速度沿着$BC→CD→DA$的路线向终点A运动.设点P运动的时间为t s,图②是点P出发t s后,$\triangle ABP$的面积S与t之间的函数关系图象.
(1)$a= $$\frac{9}{2}$,$b= $7;
(2)求线段MN的函数表达式;
由(1),得M($\frac{9}{2}$,18).因为∠A = 90°,AB = 4,AD = 6,所以当点P运动到点D时,S = $\frac{1}{2}$AB·AD = 12,所以N(7,12).设线段MN的函数表达式为S = k₁t + b₁($\frac{9}{2}$≤t≤7).把点M($\frac{9}{2}$,18),N(7,12)分别代入S = k₁t + b₁,得{$\frac{9}{2}$k₁ + b₁ = 18,7k₁ + b₁ = 12},解得{k₁ = -$\frac{12}{5}$,b₁ = $\frac{144}{5}$},所以线段MN的函数表达式为S = -$\frac{12}{5}$t + $\frac{144}{5}$($\frac{9}{2}$≤t≤7).
(3)运动几秒时,$\triangle ABP$的面积为14?
设线段OM的函数表达式为S = k₂t (0≤t≤$\frac{9}{2}$).把点M($\frac{9}{2}$,18)代入S = k₂t,得$\frac{9}{2}$k₂ = 18,解得k₂ = 4,所以线段OM的函数表达式为S = 4t(0≤t≤$\frac{9}{2}$).令S = 14,得4t = 14,解得t = $\frac{7}{2}$.在S = -$\frac{12}{5}$t + $\frac{144}{5}$中,令S = 14,得 - $\frac{12}{5}$t + $\frac{144}{5}$ = 14,解得t = $\frac{37}{6}$.综上所述，运动$\frac{7}{2}$s或$\frac{37}{6}$s时，△ABP的面积为14.

答案：
(1)$\frac{9}{2}$　7　解析:因为∠B = 90°,AD//BC,所以∠A = 180° - ∠B = 90°.观察题图可知:当点P运动到点C时,S = 18,所以$\frac{1}{2}$BC·AB = 18.因为AB = 4,所以BC = 9.因为点P的运动速度为每秒2个单位长度，所以a = 9÷2 = $\frac{9}{2}$.因为AD = 6,所以b = 10 - 6÷2 = 7.
(2)由
(1),得M($\frac{9}{2}$,18).因为∠A = 90°,AB = 4,AD = 6,所以当点P运动到点D时,S = $\frac{1}{2}$AB·AD = 12,所以N(7,12).设线段MN的函数表达式为S = k₁t + b₁($\frac{9}{2}$≤t≤7).把点M($\frac{9}{2}$,18),N(7,12)分别代入S = k₁t + b₁,得{$\frac{9}{2}$k₁ + b₁ = 18,7k₁ + b₁ = 12},解得{k₁ = -$\frac{12}{5}$,b₁ = $\frac{144}{5}$},所以线段MN的函数表达式为S = -$\frac{12}{5}$t + $\frac{144}{5}$($\frac{9}{2}$≤t≤7).
(3)设线段OM的函数表达式为S = k₂t (0≤t≤$\frac{9}{2}$).把点M($\frac{9}{2}$,18)代入S = k₂t,得$\frac{9}{2}$k₂ = 18,解得k₂ = 4,所以线段OM的函数表达式为S = 4t(0≤t≤$\frac{9}{2}$).令S = 14,得4t = 14,解得t = $\frac{7}{2}$.在S = -$\frac{12}{5}$t + $\frac{144}{5}$中,令S = 14,得 - $\frac{12}{5}$t + $\frac{144}{5}$ = 14,解得t = $\frac{37}{6}$.综上所述，运动$\frac{7}{2}$s或$\frac{37}{6}$s时，△ABP的面积为14.

26. (6分)(2025·江苏扬州期末)在一条平坦笔直的道路上依次有A,B,C三地,甲从B地骑电瓶车到C地,同时乙从B地骑摩托车到A地,到达A地后因故停留1 min,然后立即掉头(掉头时间忽略不计)按原路原速前往C地,结果乙比甲早2 min到达C地,两人均匀速运动,如图是两人距B地的路程$y(m)与时间x(min)$之间的函数图象.请解答下列问题:
(1)甲的速度为300m/min,乙的速度为800m/min;
(2)求线段FG的函数表达式;
(3)出发多少分钟后,甲、乙两人相距600 m? 请直接写出答案.
(2)由(1),得F(3,0),G(6,2400).设线段FG的函数表达式为y = kx + b(3≤x≤6).把点F(3,0),G(6,2400)分别代入y = kx + b,得{3k + b = 0,6k + b = 2400},解得{k = 800,b = -2400}.故线段FG的函数表达式为y = 800x - 2400 (3≤x≤6).
(3)$\frac{6}{11}$min或$\frac{18}{5}$min或6min

答案：
(1)300　800　解析:观察题图可知:A,B两地相距800 m,B,C两地相距2400 m,乙从B地到A地用时(3 - 1)÷2 = 1(min),所以乙的速度为800÷1 = 800(m/min),所以乙从B地到C地用时2400÷800 = 3(min),所以甲从B地到C地用时3 + 3 + 2 = 8(min),所以甲的速度为2400÷8 = 300(m/min).
(2)由
(1),得F(3,0),G(6,2400).设线段FG的函数表达式为y = kx + b(3≤x≤6).把点F(3,0),G(6,2400)分别代入y = kx + b,得{3k + b = 0,6k + b = 2400},解得{k = 800,b = -2400}.故线段FG的函数表达式为y = 800x - 2400 (3≤x≤6).
(3)当甲、乙两人相距600 m时，分类讨论如下:①若0≤x≤1,则(300 + 800)x = 600,解得x = $\frac{6}{11}$;②若2≤x≤3,则300x + 800(3 - x)=600,解得x = $\frac{18}{5}$,不合题意，舍去;③若3＜x≤6,则|300x - 800(x - 3)| = 600,解得x = $\frac{18}{5}$或6.综上所述，出发$\frac{6}{11}$min或$\frac{18}{5}$min或6min后，甲、乙两人相距600 m.