(3)设总人数为n个,80.0剟41.3?n?4?.9, ∴48?n?54, 又∵4%n为整数, ∴n?50,
即优秀的学生有52%?50?10%?260人. 24.解:(1)作MN?BO,
由垂径定理得,点N为OB的中点,
?MN?12OA,
MN?3,?OA?6,即A(?6,0),
sin?ABO?32,OA?6, ?OB?23,
即B(0,23), 设y?kx?b, 将A、B代入,
得y?33x?23, (2)NB?12OB?3,MN?3,
tan?BMN?BNMN?33, 则?BMN?30?,
??ABO?60?,
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??AMO?120?
?S阴?13?(23)2?(23)2?4??33. 3425.解:(1)由题意可得,小丽速度?设小明速度为xkm/h 由题意得,1?(16?x)?36
36?16(km/h) 2.25?x?20
答:小明的速度为20km/h,小丽的速度为16km/h. (2)由图象可得,点E表示小明到了甲地,此时小丽没到,
?点E的横坐标?369?, 2059144点E的纵坐标??16?,
55?点E(,
95144) 526.解:(1)如图1,连接AE并延长交圆E于点C,作AC的中垂线交圆于点B,D,四边形ABCD即为所求.
(2)①如图2,连接AC,BD交于点O,连接EB交AC于点G,连接DG并延长交CB 于点F,F即为所求
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②如图3所示,AH即为所求.
27.解:(1)令x?0,则y??4, ?C(0,?4),
OA?OB,
?对称轴在y轴右侧,即?b2a?0 a?0,?b?0;
(2)①过点D作DM?Oy,
则
DCDMMCCA?OA?CO?12,
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?DM?12AO, 设A(?2m,0)m?0, 则AO?2m,DM?m
OC?4,?CM?2,
?D(m,?6),B(4m,0),
则
MDBO?MEOE?6OE?OE, ?OE?8,
SBEF?12?4?4m?8, ?m?1,
?A(?2,0),B(4,0),
设y?a(x?2)(x?4),
即y?ax2?2ax?8a, 令x?0,则y??8a, ?C(0,?8a),
??8a??4,a?12, ?y?12x2?x?4; ②由①知B(4m,0)C(0,?4)D(m,?6), 则?CBD一定为锐角,
CB2?16m2?16,CD2?m2?4,DB2?9m2?36,19
当?CDB为锐角时, CD2?DB2?CB2,
m2?4?9m2?36?16m2?16,
解得,?2?m?2; 当?BCD为锐角时, CD2?CB2?DB2,
m2?4?16m2?16?9m2?36,
解得,m?2或m??2?舍?,
综上,2?m?2,22?2m?4;
故22?OA?4. 28.解:(1)①如图1中,
四边形ABCD是矩形,
??ABC?90?,
?AC?AB2?BC2?21,
?PCB???ACB,?PB?C??ABC?90?, ?PCB?∽ACB,
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