数学试卷
GB?BD?cos30°?100?3·········································································· 6分 ?503． ·
2在Rt△ADG中，?GDA?105°?60°?45? ··································································· 7分 ∴GD?GA?50， ·············································································································· 8分 ∴AB?AG?GB?50?503（米） ··············································································· 9分 答：索道长50?503米． ········································································································ ·············································································································································· 10分 25. （1）描点略． ·············································································································· 1分 设y?kx?b，用任两点代入求得y??100x?5000， ·················································· 3分 再用另两点代入解析式验证． ···························································································· 4分 （2）?y?z，??100x?5000?400x，
···························································································································· 6分 ?x?10． ·
······································································· 7分 ?总销售收入?10?4000?40000（元） ·
?农副产品的市场价格是10元/千克，
农民的总销售收入是40000元． ························································································ 8分 （3）设这时该农副产品的市场价格为a元/千克，
则a(?100a?5000)?40000?17600， ·········································································· 10分 解之得：a1?18，a2?32．
································································································ 11分 ?0?a?30，?a?18． ·
······························································ 12分 ?这时该农副产品的市场价格为18元/千克． ·
?， AC?CD26. （1）证明：∵C是?AD的中点，∴?∴∠CAD=∠ABC???????????????????1分 ∵AB是⊙O的直径，∴∠ACB=90°。 ∴∠CAD+∠AQC=90°
又CE⊥AB，∴∠ABC+∠PCQ=90°
∴∠AQC=∠PCQ???????????????????2分 ∴在△PCQ中，PC=PQ，
AC??AE ∵CE⊥直径AB，∴?? AE?CD∴?∴∠CAD=∠ACE。
∴在△APC中，有PA=PC，???????????????????3分 ∴PA=PC=PQ
∴P是△ACQ的外心。???????????????????4分
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数学试卷
（2）解：∵CE⊥直径AB于F， ∴在Rt△BCF中，由tan∠ABC=得BF?CF3?，CF=8， BF4432CF?。???????????????????5分 3340???????????????????6分 3∴由勾股定理，得BC?CF2?BF2?∵AB是⊙O的直径，
∴在Rt△ACB中，由tan∠ABC=
得AC?AC340 ?，BC?BC433BC?10。???????????????????7分 4易知Rt△ACB∽Rt△QCA，∴AC2?CQ?BC
AC215∴CQ??。???????????????????8分
BC2（3）证明：∵AB是⊙O的直径，∴∠ACB=90°
∴∠DAB+∠ABD=90°
又CF⊥AB，∴∠ABG+∠G=90° ∴∠DAB=∠G；
∴Rt△AFP∽Rt△GFB，???????????????????10分 ∴
AFFP，即AF?BF?FP?FG ?FGBF易知Rt△ACF∽Rt△CBF，
∴FG2?AF?BF（或由射影定理得）
∴FC2?PF?FG???????????????????11分 由（1），知PC=PQ，∴FP+PQ=FP+PC=FC
∴(FP?PQ)2?FP?FG。???????????????????12分
27（1）C(3,2),D(1,3)；???????????????????2分
（2）设抛物线为y?ax2?bx?c，抛物线过(0,1),(3,2),(1,3)，
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数学试卷
5?a??,?6?c?1,?17??,???????????????????4分 ?a?b?c?3,解得?b?6?9a?3b?c?2.???c?1.??∴y??5217x?x?1．???????????????????????5分 66（3）①当点A运动到点F时，t?1,
当0?t?1时，如图1，
∵?OFA??GFB'， tan?OFA?∴tan?GFB'?∴S?FB'G?OA1?, OF2GB'GB'15??,∴GB'?t,
2FB'25t图1
115t52FB'?GB'??5t??t；??8分 2224 ②当点C运动到x轴上时，t?2，
当1?t?2时，如图2，
A'B'?AB?22?12?5,
∴A'F?5t?5,∴A'G?图2
5t?5， 2∵B'H?5t， 2∴S梯形A'B'HG?（A'G?B'H)?A'B' ?1215t?55t(?)?5 222?
图3
55t?；????（10分） 24③当点D运动到x轴上时，t?3， 当2?t?3时，如图3，
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数学试卷
∵A'G?5t?5， 25t?535?5t， ?22∴GD'?5?∵S?AOF?1?1?2?1,OA?1， 2图4
?AOF∽?GD'H SGD'2)， ∴?GD'H?(S?AOFOA∴S?GD'H?(35?5t2)， 235?5t2) 22∴S五边形GA'B'C'H?（5）?（ =?521525t?t?．???（12分） 424（解法不同的按踩分点给分）
∵t?3,BB'?AA'?35,
∴S阴影?S矩形BB'C'C?S矩形AA'D'D ??????????????????（13分） =AD?AA'
=5?35?15．???????????????????????（14分）
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