2007—2014年陇南市中考题（含答案）(10)

∵ 四边形ABCD是平行四边形，

∴ AD∥BF，∴ ∠D=∠ECF． ··························································· 3分

∵ E是CD的中点，∴ DE = CE．

······························· 1分

相切： ······························· 3分

相交： ······························· 5分

其它：

························································ 6分

21． 本小题满分8分

解：（1）7，158． ·········································································· 4分

（2）设所求的解析式为y?kx?b， ·························································· 5分 ∵ 点（0，15）、（1，7）在图像上，

∴ ??15?b,?7?k?b. ……………………………………………………………………… 6分 解得 k??8，b?15．

∴ 所求的解析式为y??8x?15． (0≤x≤158) …………………………… 8分 说明：只要求对k??8、b?15，不写最后一步，或者未注明x的取值范围，都不扣分．22． 本小题满分8分

证明：(1)

又 ∠AED=∠FEC， ··································· 4分

∴ △ADE≌△FCE． ································· 5分 (2) D．或填“平行四边形”． ······························ 8分 23． 本小题满分10分

解；（1）不及格，及格； ········································································ 4分 （2）抽到的考生培训后的及格与优秀率为（16+8）÷32=75%， ······················ 6分 由此，可以估计八年级320名学生培训后的及格与优秀率为75%． ················ 8分 所以，八年级320名学生培训后的及格与优秀人数为75%×320=240． ·············10分 四、解答题(二)：本大题共5小题，共50分． 24． 本小题满分8分

解：(1)56； ···························································································· 3分 （2）如图，△O1 O2 O3是边长为8mm的正三角形， 作底边O2O3上的高O1 D． ······························· 4分

则 O1D=O1O3·sin60°=43≈6.92． ··················· 6分 O1 ∴ AD=2(O1D+4)=2×10.92≈21.8（mm）． ··············· 8分 说明：(1)用勾股定理求O1D，参考本标准评分； O2 D O3 （2）在如图大正三角形中求高后再求AD，

也参考本标准评分．

25． 本小题满分10分

解：设花边的宽为x分米， ······························································· 1分 根据题意，得(2x?6)(2x?3)?40． ············································· 5分

解得x，x111?12??4． ··························································· 8分 x112=?4不合题意，舍去． ································································ 9分 答: 花边的宽为1米． ·····························································10分 说明：不答不扣分． 26． 本小题满分10分

解：（1）如图①，作DE⊥BC于E， ·················· 1分

图①

∵ AD∥BC，∠B=90°， 28． 本小题满分12分

∴ ∠A=90°．又∠DEB=90°，

∴ 四边形ABED是矩形． ·································································· 2分 ∴ BE=AD=2， ∴ EC=BC-BE=3． ······················································· 3分 在Rt△DEC中，DE= EC·tanC =3?43=4． ·········································· 5分 （2）如图②，作BF⊥CD于F． ·························································· 6分 方法一：