《2025年学霸甘肃少年儿童出版社八年级数学上册苏科版》

21.(8分)长清的园博园广场视野开阔,阻挡物少,成为不少市民放风筝的最佳场所,某校八年级(1)班的小明和小亮学习了“勾股定理”之后,为了测得风筝的垂直高度$CE$,他们进行了如下操作：①测得水平距离$BD$的长为15米；②根据手中剩余线的长度计算出风筝线$BC$的长为25米；③牵线放风筝的小明的身高为1.6米.
(1)求风筝的垂直高度$CE$；
(2)如果小明想风筝沿$CD$方向下降12米,那么他应该往回收线多少米？

答案：

(1)在Rt△CDB中,由勾股定理,得$CD^{2}=BC^{2}-BD^{2}=25^{2}-15^{2}=400$,
∴CD = 20米,
∴CE = CD + DE = 20 + 1.6 = 21.6(米).
答:风筝的垂直高度CE为21.6米
(2)如图,由题意得,CM = 12米,
∴DM = 8米,

∴$BM^{2}=DM^{2}+BD^{2}=8^{2}+15^{2}=289$,
∴BM = 17米,
∴BC - BM = 25 - 17 = 8(米).
答:小明应该往回收线8米

22.(8分)如图,在$\triangle ABC$中,$\angle ABC = 45^{\circ}$,$CD\perp AB$,$BE\perp AC$,垂足分别为$D$,$E$,$F为BC$的中点,$BE与DF$,$DC分别交于点G$,$H$,$\angle ABE= \angle CBE$.
(1)线段$BH与AC$相等吗？若相等,给予证明；若不相等,请说明理由.
(2)求证：$BG^2 - GE^2 = EA^2$.

答案：
(1)BH = AC. 证明:
∵CD⊥AB,BE⊥AC,
∴∠BDH = ∠BEA = ∠CDA = 90°.
∵∠ABC = 45°,
∴∠BCD = 180° - 90° - 45° = 45° = ∠ABC,
∴DB = DC.
∵∠BDH = ∠BEA = ∠CDA = 90°,
∴∠A + ∠ACD = 90°,∠A + ∠HBD = 90°,
∴∠HBD = ∠ACD.
∵在△DBH和△DCA中,$\left\{\begin{array}{l} ∠BDH = ∠CDA,\\ BD = CD,\\ ∠HBD = ∠ACD,\end{array}\right.$
∴△DBH≌△DCA(ASA),
∴BH = AC.
(2)连接CG,由
(1)知,DB = CD.
∵F为BC的中点,
∴DF垂直平分BC,
∴BG = CG.
∵∠ABE = ∠CBE,BE⊥AC,
∴EC = EA.在Rt△CGE中,由勾股定理,得$CG^{2}-GE^{2}=CE^{2}$,
∴$BG^{2}-GE^{2}=EA^{2}$.

2

3.(9分)已知：$\triangle ABC$(如图所示).
(1)请在图中用无刻度的直尺和圆规作图：求作一点$D$,使得点$D到边AB$,$AC$的距离相等,且满足$DB = DC$.(不写作法,保留作图痕迹)
(2)若$AB = 15$,$AC = 9$,在(1)的基础上,过点$D作DE\perp AB于E$,求$BE$的长.(如需画草图,请使用备用图)

答案：

(1)如图①所示. 解析:在AB,AC上分别截取AN,AM,使AN = AM;分别以点M和点N为圆心,适当长(大于$\frac{1}{2}MN$的长)为半径作圆弧,在∠BAC内,两弧交于点O;连接AO并延长,分别以点B和点C为圆心,以大于$\frac{1}{2}BC$的长度为半径作弧,两弧相交于点Q,P;作直线PQ,直线PQ与射线AO交于点D,点D即为所求.

(2)如图②,过点D作DE⊥AB交AB于点E,过点D作DF⊥AC交AC的延长线于点F.由
(1)知BD = DC,∠BAD = ∠CAD,
∴DE = DF.在Rt△BDE与Rt△CDF中,$\left\{\begin{array}{l} BD = CD,\\ DE = DF,\end{array}\right.$
∴Rt△BDE≌Rt△CDF(HL),
∴BE = CF.在Rt△ADE与Rt△ADF中,$\left\{\begin{array}{l} AD = AD,\\ DE = DF,\end{array}\right.$
∴Rt△ADE≌Rt△ADF(HL),
∴AE = AF,
∴BE = CF = AB - AE = AB - (AC + CF),即BE = AB - AC - BE,
∴$BE = \frac{AB - AC}{2}$.
∵AB = 15,AC = 9,
∴$BE = \frac{15 - 9}{2}=3$.

24.(9分)如图,$C为线段BD$上一动点,分别过点$B$,$D作AB\perp BD$,$ED\perp BD$,连接$AC$,$EC$,已知$AB = 5$,$DE = 1$,$BD = 8$,设$CD = x$.
(1)用含$x的代数式表示AC + CE$的长；
(2)请问点$C$满足什么条件时,$AC + CE$的值最小？
(3)根据(2)中的规律和结论,请构图求出代数式$\sqrt{x^2 + 4}+\sqrt{(12 - x)^2 + 9}$的最小值.

答案：

(1)根据题意,AB⊥BD,ED⊥BD,AB = 5,DE = 1,BD = 8,设CD = x,则BC = BD - CD = 8 - x,利用勾股定理,得$AC = \sqrt{AB^{2}+BC^{2}}=\sqrt{5^{2}+(8 - x)^{2}}=\sqrt{25 + (8 - x)^{2}}$,$CE = \sqrt{CD^{2}+DE^{2}}=\sqrt{x^{2}+1}$,
∴$AC + CE = \sqrt{25 + (8 - x)^{2}}+\sqrt{x^{2}+1}$.
(2)根据题意,连接AE,则AC + CE ≥ AE,当A,C,E三点共线时,AC + CE的值最小.
(3)根据$\sqrt{x^{2}+4}+\sqrt{(12 - x)^{2}+9}$,构造AB = 3,DE = 2,BD = 12,CD = x.如图所示,当A,C,E三点共线时,AC + CE最小,延长ED到点F,过点A作AF⊥DF于点F,易得AF = BD = 12,AB = DF = 3,EF = ED + DF = 5.故$AE = \sqrt{AF^{2}+EF^{2}}=13$.