∵∠2 = 45°,∴BE = BD,∠EBD = 90°.∴AC = BD. 延长AC交DB的延长线于F,如图4.∵BE∥AC,∴∠AFD = 90°.∴AC⊥BD.
(3)如图5,过点B作BE∥CA交DO于E,∴∠BEO = ∠ACO.
又∵∠BOE = ∠AOC , ∴△BOE ∽ △AOC.
D M 2 E A N
1 ∴
BEAC?BOAO.
又∵OB = kAO,
B
O C
图5
由(2)的方法易得 BE = BD.∴BDAC?k.
25.解:(1)y = 2t;(2)当BP = 1时,有两种情形:
①如图6,若点P从点M向点B运动,有 MB = BC= 4,MP = MQ = 3,
21A E ∴PQ = 6.连接EM,
D ∵△EPQ是等边三角形,∴EM⊥PQ.∴EM?33. ∵AB = 33,∴点E在AD上.
B P
M 图6
Q C
∴△EPQ与梯形ABCD重叠部分就是△EPQ,其面
积为93.
②若点P从点B向点M运动,由题意得 t?5.
PQ = BM + MQ?BP = 8,PC = 7.设PE与AD交于点F,QE与AD或AD的
E A H F G D 延长线交于点G,过点P作PH⊥AD于点H,则 HP = 33,AH = 1.在Rt△HPF中,∠HPF = 30°, ∴HF = 3,PF = 6.∴FG = FE = 2.又∵FD = 2, ∴点G与点D重合,如图7.此时△EPQ与梯形ABCD
B P M 图7
C Q
的重叠部分就是梯形FPCG,其面积为
11
2723.
(3)能.4≤t≤5.
26.解:(1)140 57500;
(2)w内 = x(y -20)- 62500 = ?w外 = ?(3)当x = ?11002
1100x2+130 x?62500,
x+(150?a)x.
1302?(?1100= 6500时,w内最大;分
)124?(?)?(?62500)?1300?(150?a)100由题意得 , ?114?(?)4?(?)1001002解得a1 = 30,a2 = 270(不合题意,舍去).所以 a = 30. (4)当x = 5000时,w内 = 337500, w外 =?. 5000a?500000若w内 < w外,则a<32.5; 若w内 = w外,则a = 32.5; 若w内 > w外,则a>32.5.
所以,当10≤ a <32.5时,选择在国外销售;
当a = 32.5时,在国外和国内销售都一样;
当32.5< a ≤40时,选择在国内销售.
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2010年河北省中考数学试题及答案
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