广东省广州花都区2018-2019学年八年级下学期期中考试数学试题（扫描版含答案）

2018学年第二学期八年级数学期中抽测参考答案 一、选择题： 1、 2、 3、 4、 5、 6、 7、 8、 9、 10、 A D D C C C C A B A 二、填空题： 11、 12、 13、 14、 15、 16、 12 18 AB=BC (多个答案) 5 11 三、解答题： 17、（1） 解：原式= ``````````````````````````````4 分 = ``````````````````````````````5分 （2） 解：原式=``````````````````````````````2分 =``````````````````````````````3 分 =``````````````````````````````5分 （3） 解：原式= `````````````````3 分 = ``````````````````````````````5 分 （4） 解：原式=```````````````````````3 分 =``````````````````````````````4 分 = -13-``````````````````````````````5分 18、方法一：（直接代入） 解：当时， 原式 ``````````````````````````````2 分 ``````````````````````````````7 分 =0````````````````````````````````````````````9 分 方法二：（先变形） 解： ````````````5分 当时， ````````````````6 分 ```````````````````7分 =2-2 `````````````````````8分 =0 ````````````````````9分 19、解： ``````````````````````````````3分 =3 ``````````````````````````````5 分 ``````````````````````````````9分 20、解： ``````````````````````````````4 分 四边形ABCD的周长= ````````6分 ````````7分 四边形ABCD的面积是: ````````10分 21、 证明：在菱形ABCD中，AB=AD=BC=CD, ∠B=∠D, ``````````````````3分 又E、F分别是BC、CD的中点. BE=DF ``````````````````````````````5 分 在△ABE和△ADF中 ````````````````````````````8 分 △ABE△ADF(SAS) ``````````````````````````````9分 AE=AF ``````````````````````````````10分 22、解：在矩形ABCD中， ∠ABC=90°,AC=BD，AO=AC, BO=BD， ````````3分 AO=BO ``````````````````````````````4分 ∠AOD=120° ∠AOB=60° △AOB是等边三角形```````````````````````6分 AO=BO=AB=3， AC=2AO=6 ``````````````````````````````7分 在Rt△ABC中，由勾股定理得： ````9分 矩形的面积=ABBC=``````````````````````````````10分 23、（1）证明：DE∥AC ，DF∥AB 四边形AEDF是平行四边形``````````````````````````````2分 DE∥AC ∠EDA=∠DAF``````````````````````````````3分 AD是△ABC的角平分线 ∠EAD=∠DAF``````````````````````````````4分 ∠EDA=∠EAD EA=ED``````````````````````````````6分 平行四边形AEDF是菱形```````````````7分 （2）当△ABC满足∠BAC=90°时，四边形AEDF是正方形。 理由如下： 四边形AEDF是菱形且∠BAC=90° 四边形AEDF是正方形（有一个内角是直角的菱形是正方形）````````10分 24、（1）AP=t， BQ=26-3t ``````````````````````````````2分 （2）由题意可得：PD=AD-AP=24-t， `````3分 QC=3t```````````4分 AD∥BC PD∥QC 设当运动时间为t秒时PD=QC，此时四边形PQCD为平行四边形。 由PD=QC得，24-t=3t ``````````````````````````6分 解得t=6 当运动时间为6秒时，四边形PQCD为平行四边形。```````7分 （3） AD∥BC AP∥BQ 设当运动时间为t秒时AP=BQ，四边形ABQP为平行四边形。 由AP=BQ得：t=26-3t ``````````````````````````````9分 解得：t=``````````````````````````````10分 又∠B=90° 平行四边形ABQP为矩形。 当运动时间为秒时，四边形ABQP为矩形。`````````12分 25、（1）∠AMB=2∠MAE ``````````````````````````````1分 理由：四边形ABCD是正方形 AD∥BC ∠DAM=∠AMB AE平分∠DAM ∠MAE=∠DAE=∠DAM ∠DAM=2∠MAE ∠AMB=2∠MAE `````````````````````````````2分 (2)过点E作EFAM于点F,连接EM `````````````````3分 四边形ABCD是正 ... ...