《2025年轻松作业本八年级数学上册苏科版》

8. 如图,一个无盖的长方体盒子的长、宽、高分别为3.5cm,3.5cm,24cm,一只蚂蚁想从盒底的点A沿盒的表面爬到盒顶的点B,则它爬行的最短路程是______25cm.

答案： 25

9. 被誉为“中国数学界的图腾”的“赵爽弦图”,是用四个全等的直角三角形拼成如图①所示的大正方形,中间也是一个正方形,其中四个直角三角形的直角边长分别为a,b($a ＜ b$),斜边长为c.若将这四个全等的直角三角形无缝隙无重叠地拼接成如图②,得到图形ABCDEFGH.若该图形的周长为48,$OH = 6$,则$b = $8,$c = $10.

答案： 8 10

11. 如图,某沿海城市A接到台风预警,在该市正南方向340km的B处有一台风中心,沿BD方向以15km/h的速度移动,已知城市A到BC的距离AD为160km.
(1) 台风中心经过多长时间从B点移到D点?
(2) 如果在距台风中心200km的圆形区域内都将受到台风的影响,那么A市受到台风影响的时间持续多少小时?

答案：
解：
(1)$AD⊥BC$，$AB = 340km$，$AD = 160km$，$BD = \sqrt{AB^{2} - AD^{2}} = 300(km)$，$\because 300÷15 = 20(h)$，$\therefore$台风中心经过20h从B点移到D点；
(2)在射线BC上取点E、F，使得$AE = AF = 200km$，由$AD⊥BC$得$DE = DF$，在$Rt△AED$中，$ED = \sqrt{AE^{2} - AD^{2}} = 120(km)$，$\therefore EF = 2ED = 240km$，$\therefore t = 240÷15 = 16(h)$，$\therefore$时间持续16h。

12. (2024秋·扬州广陵区期中)在$\triangle ABC$中,$AB = AC = 10$,$BC = 16$.点D是BC的中点,点E是线段BD上的动点,过点E作$EF\perp BD$,交AB于点F.连接AE,若$\angle AEF = \angle B$.
(1) 求证:$AE\perp AC$;
(2) 求DE的长.

(1)证明：$\because AB = AC$，$\therefore ∠B = ∠C$。$\because EF⊥BD$，$\therefore ∠AEF + ∠AED = 90^{\circ}$。$\because ∠AEF = ∠B$，$∠B = ∠C$，$\therefore ∠C + ∠AED = 90^{\circ}$，$\therefore ∠EAC = 90^{\circ}$，$\therefore AE⊥AC$；
(2)解：$\because ∠EAC = 90^{\circ}$，$\therefore AE^{2} + AC^{2} = CE^{2}$。$\because CE = CD + DE = DE + 8$，$\therefore AE^{2} = CE^{2} - AC^{2} = (DE + 8)^{2} - 10^{2}$。
$\because AB = AC$，点D是BC的中点，$\therefore BD = DC = \frac{1}{2}×16 = 8$，$BC = 16$，$AD⊥BC$，$\therefore AD = \sqrt{AC^{2} - CD^{2}} = \sqrt{10^{2} - 8^{2}} = 6$。在$Rt△ADE$中，$AE^{2} = AD^{2} + DE^{2} = 6^{2} + DE^{2}$，$\therefore (DE + 8)^{2} - 10^{2} = 6^{2} + DE^{2}$，解得：$DE = $4.5。

答案：
(1)证明：$\because AB = AC$，$\therefore ∠B = ∠C$。$\because EF⊥BD$，$\therefore ∠AEF + ∠AED = 90^{\circ}$。$\because ∠AEF = ∠B$，$∠B = ∠C$，$\therefore ∠C + ∠AED = 90^{\circ}$，$\therefore ∠EAC = 90^{\circ}$，$\therefore AE⊥AC$；
(2)解：$\because ∠EAC = 90^{\circ}$，$\therefore AE^{2} + AC^{2} = CE^{2}$。$\because CE = CD + DE = DE + 8$，$\therefore AE^{2} = CE^{2} - AC^{2} = (DE + 8)^{2} - 10^{2}$。
$\because AB = AC$，点D是BC的中点，$\therefore BD = DC = \frac{1}{2}×16 = 8$，$BC = 16$，$AD⊥BC$，$\therefore AD = \sqrt{AC^{2} - CD^{2}} = \sqrt{10^{2} - 8^{2}} = 6$。在$Rt△ADE$中，$AE^{2} = AD^{2} + DE^{2} = 6^{2} + DE^{2}$，$\therefore (DE + 8)^{2} - 10^{2} = 6^{2} + DE^{2}$，解得：$DE = 4.5$。