www.qhpta.com

来源:爱牙日 时间:2018-10-04 11:00:07 阅读:

【www.zhuodaoren.com--爱牙日】

www.qhpta.com(共10篇)

www.qhpta.com(一):

已知:如图:菱形ABCD中,∠BAD=120°,动点P在直线BC上运动,作∠APM=60°,且直线PM与直线CD相交于点Q,Q点到直线BC的距离为QH.

(1)若P在线段BC上运动,求证CP=DQ;
(2)若P在线段BC上运动,探求线段AC、CP、CH的一个数量关系,并证明你的结论;
(3)若动点P在直线BC上运动,菱形ABCD周长为8,AQ=
6

(1)证明:作PE∥CD交AC于E,则△CPE是等边三角形∠EPQ=∠CQP.
又∵∠APE+∠EPQ=60°,∠CQP+∠CPQ=60°
∴∠APE=∠CPQ
又∵∠AEP=∠QCP=120°,PE=PC
∴△APE≌△QPC
∴AE=QC,AP=PQ,
∴△APQ是等边三角形,
∴∠2+∠3=60°,
∵∠1+∠2=60°,
∴∠1=∠3,
在△AQD和△APC中

∠D=∠ACP
∠1=∠3
AQ=AP

∴△AQD≌△APC(AAS),
∴CP=DQ.

(2)∵AC=CD,CD=CQ+QD,
∴AC=CQ+QD,
∵CP=DQ,
∴AC=CQ+PC,
又∵∠CHQ=90°,∠QCH=60°,
∴∠CQH=30°,
∴CQ=2CH,
∴AC=CP+2CH;

(3)此题分两种情况讨论:
①当点P在射线BC上时;
设CH=x,则QH=
3
x,PC=2-2x,由勾股定理得,
3
x)2+(2-x)2=6,解得x=
3
2
(舍去负的),
x=
1+
3
2
,∴QH=
3
x=

www.qhpta.com(二):

(2014•乐清市二模)如图,在直角坐标系xOy中,△ABC的顶点坐标为A(-8,0),B(3,0),C(0,4).动点P从点A出发,沿着AB以每秒1个单位长度的速度向终点B运动;动点Q从点B出发,沿着射线BC,以每秒1个单位长度的速度运动,当点P到达B时,点Q也停止运动.P,Q两点同时开始运动,设运动时间为t秒.
(1)求线段BC的长度;
(2)当△APQ为等腰三角形时,求t的值;
(3)设△APQ的外接圆的圆心为M,当点C在⊙M上时,请求出t的值.

(1)∵B(3,0),C(0,4),
∴OB=3,OC=4.
在Rt△BOC中,
∵∠BOC=90°,
∴BC=

OB2+OC2
=5.

(2)①当PA=PQ时,作QH⊥x轴,垂足为H,如图1,

由题可得:PQ=PA=BQ=t.
∵A(-8,0),B(3,0),
∴OA=8,OB=3,∴AB=11.
∵sin∠HBQ=
QH
BQ
=
OC
BC
=
4
5
,cos∠HBQ=
BH
BQ
=
OB
BC
=
3
5

∴QH=
4
5
t,BH=
3
5
t.
∴PH=
.
11−t−
3
5
t
.
=
.
11−
8
5
t
.

在Rt△PHQ中,
∵QH2+PH2=PQ2
(
4
5
t)2+(11−
8
5
t)2t2

解得:t1=5,t2=11.
②当QP=QA时,作QH⊥x轴,垂足为H,如图2,

则AH=PH=
1
2
AP=
1
2
t,BH=11−

www.qhpta.com(三):

想问一下大家有没有在什麽教辅书上看到这篇文章,或是在哪年的中考有这篇阅读理解!谢
can a simple idea help make the world a better place?each week we ask a gust to give an idea to improve all our lives.Here,linguist Daniel Everett says we should all live a weekwith complete strangers.The following is what he said:
下面的不写了,这是第一段,
我怀疑这是哪本书上P57页的东西
如果可以的话帮我查一下这是哪本书

Can a Green Coupon Book Fundraiser Create Green Consumers?
LOS ANGELES, Sept. 19 /PRNewswire/ -- Today, everyone knows it""s cool to be green. It""s in the papers everyday. So why hasn""t the idea of "Green Consumerism" taken hold among middle class Americans?
It""s simple: convenience and price.
An environmentally friendly home improvement store isn""t going to just appear before our eyes. And try finding any selection of "fair trade" clothing or "cruelty free" cosmetics in a department store. Retailers are slow to move towards products that traditionally have lower profit margins.
But someone is out to change all of this. PTA mom, environmental activist and business owner, Jennifer Caronna. She started Fundraising Green with a focus on healthy kids, healthy homes and a healthy planet. Her new fundraising product is a Green Coupon Book. It""s called myGreenSpark Redemption Book and will be available this November. It""s full of coupon offers from Green, Ethical and Healthy businesses that serve Southern California, the U.S. and the world.
When asked to explain, Jennifer had this to say, "Our Green Coupon Book serves many purposes. First, it raises money for our customers in an ethical manner, while promoting Green and Healthy habits. Second, instead of participating in school assemblies that teach them how to shill cookie dough to family and friends, students are exposed to the idea of living green and eating healthy and how both are equally as important in creating healthy communities. Further to this point, our redemption book doubles as an educational tool, with explanations on the stubs of each coupon telling why we as consumers should use this product or service to better our health and the environment. We think that our book will provide the "Green Spark" that families need to see just how easy all of this can be. And finally, when used, the coupons do create "Green Consumers". The more coupons we redeem, the less impact we have on our environment and the healthier we become. We encourage families to explore the contents of the book together. To find a few new products and services, as a family, that will lead them on the path to a Greener, Healthier, Happier lifestyle."
Jennifer is also quick to point out that besides a lot of fun and healthy kids activities, the myGreenSpark Redemption Book is loaded with coupons for everything you need to Green your life. Green your home, your clothes, your kids, your pets, your food and even your vacation. And all with exclusive myGreenSpark discounts. "It""s been so rewarding putting this book together. Business Partners like REI, The Aquarium of the Pacific, Flexcar, Bikestation, No Sweat Apparel, 118 Degrees, TheVegitarianSite.com and Organic Bouquet have been very supportive and are excited about our project."
Jennifer continues, "As the dust settles, we""re seeing that a significant number of our coupon offers are web-based. We""ve come to the conclusion that the Internet offers a very comprehensive and cost effective way to Green the life of most American families. And just think, you never even have to get in your car."
"This gave us an idea. Reaction from around the country has been great. People keep signing up on our website asking us when we""ll have a Green Coupon Book for their area. We""re working on more cities as we speak, but in the meantime, we""ve decided to produce an "Internet Only" version of the myGreenSpark Redemption Book and offer it as a new nationwide eco-friendly fundraiser. At the very least we""ll educate a ton of people. At best we""ll create Green consumers, influence big business to go Green and help in the fight to create a sustainable economy."
The myGreenSpark Redemption Book Fundraiser will be available to schools, clubs, non-profits and foundations this November. It""s a zero administration, web-based fundraiser. Customers never collect any funds or distribute any product, but receive 40% - 50% of the proceeds.
To learn more about Fundraising Green visit http://www.FundraisingGreen.com or email info@FundraisingGreen.com. Green teens and Green mentors please visit http://www.BeCoolBeGreen.com and learn how you can get involved to make a difference.
Source: Fundraising Green

www.qhpta.com(四):

英语翻译
1.Table tennis player WangTao won his first-ever men"s singles World title on October 14.这句中的first-ever 的准确翻译?
2.Most students in the US don"t wear school uniforms.And they don"t have group exercises each day as the way you do.这句中的group exercises 的准确翻译?
【www.qhpta.com】

10月14日,乒乓球选手王陶赢得了有史以来他的第一个男子单打世界冠军 first-ever 前所未有,有史以来
大多数美国在校生是不穿校服的.而且他们不像你们那样要做集体锻炼(如广播体操).group exercises 集体操,团体操

www.qhpta.com(五):

如图,以Rt△ABO的直角顶点O为原点,OA所在的直线为x轴,OB所在的直线为y轴,建立平面直角坐标系.已知OA=4,OB=3,一动点P从O出发沿OA方向,以每秒1个单位长度的速度向A点匀速运动,到达A点后立即以原速沿AO返回;点Q从A点出发沿AB以每秒1个单位长度的速度向点B匀速运动.当Q到达B时,P、Q两点同时停止运动,设P、Q运动的时间为t秒(t>0).
(1)试求出△APQ的面积S与运动时间t之间的函数关系式;
(2)在某一时刻将△APQ沿着PQ翻折,使得点A恰好落在AB边的点D处,如图①.求出此时△APQ的面积.
(3)在点P从O向A运动的过程中,在y轴上是否存在着点E使得四边形PQBE为等腰梯形?若存在,求出点E的坐标;若不存在,请说明理由.
(4)伴随着P、Q两点的运动,线段PQ的垂直平分线DF交PQ于点D,交折线QB-BO-OP于点F. 当DF经过原点O时,请直接写出t的值.

【www.qhpta.com】

(1)在Rt△AOB中,OA=4,OB=3
∴AB=

42+32
=5
①P由O向A运动时,OP=AQ=t,AP=4-t
过Q作QH⊥AP于H点.
由QH∥BO,得
QH
AQ
OB
AB
,得QH=
3
5
t

S△APQ
1
2
AP•QH=
1
2
(4−t)•
3
5
t

S△APQ=−
3
10
t2+
6
5
t
(0<t<4)
②当4<t≤5时,即P由A向O运动时,AP=t-4AQ=t
sin∠BAO=
QH
t
3
5

QH=
3
5
t

s△APQ=
1
2
(t−4)•
3
5
t

=
3
10
t2
6
5
t

综上所述,S△APQ=
3
10
t2+
6
5
t(0<t<4)
3
10
t2
6
5
t(4<t≤5)


(2)由题意知,此时△APQ≌△DPQ,∠AQP=90°,
∴cosA=<

www.qhpta.com(六):

如图,在直角梯形OABC中,OA∥CB,A、B两点的坐标分别为A(15,0),B(10,12),动点P、Q分别从O、B两点出发,点P以每秒2个单位的速度沿OA向终点A运动,点Q以每秒1个单位的速度沿BC向C运动,当点P停止运动时,点Q也同时停止运动.线段OB、PQ相交于点D,过点D作DE∥OA,交AB于点E,射线QE交x轴于点F.设动点PQ运作业帮动时间为t(单位:秒).
(1)当t为何值时,四边形PABQ是等腰梯形,请写出推理过程;
(2)当t=2秒时,求梯形OFBC的面积;
(3)当t为何值时,△PQF是等腰三角形?请写出推理过程.

(1)如图,过B作BG⊥OA于G,
则AB=

AQ
AP
BG2+GA2
122+(15−10)2
169
=13.
过Q作QH⊥OA于H,
则QP=
QH2+PH2
122+(10−t−2t)2
144+(10−3t)2

要使四边形PABQ是等腰梯形,则AB=QP,
144+(10−3t)2
=13

∴t=
5
3
,或t=5(此时PABQ是平行四边形,不合题意,舍去);
∴t=
5
3

(2)当t=2时,OP=4,CQ=10-2=8,QB=2.
∵CB∥DE∥OF,

www.qhpta.com(七):

(2008•哈尔滨)在矩形ABCD中,点E是AD边上一点,连接BE,且∠ABE=30°,BE=DE,连接BD.点P从点E出发沿射线ED运动,过点P作PQ∥BD交直线BE于点Q.
(1)当点P在线段ED上时(如图1),求证:BE=PD+
3
3
【www.qhpta.com】

(1)证明:∵∠A=90°∠ABE=30°,
∴∠AEB=60°.
∵EB=ED,
∴∠EBD=∠EDB=30°.
∵PQ∥BD,
∴∠EQP=∠EBD.
∠EPQ=∠EDB.
∴∠EPQ=∠EQP=30°,
∴EQ=EP.                                     
过点E作EM⊥QP垂足为M.则PQ=2PM.
∵∠EPM=30°,∴PM=

3
2
PE,PE=
3
3
PQ.             
∵BE=DE=PD+PE,
∴BE=PD+
3
3
PQ.                               

(2)由题意知AE=
1
2
BE,
∴DE=BE=2AE.
∵AD=BC=6,
∴2AE=DE=BE=4.                              
当点P在线段ED上时(如图1),
过点Q做QH⊥AD于点H,则QH=
1
2
PQ=
1
2
x.
由(1)得PD=BE-
3
3
x,PD=4-
3
3
x.
∴y=
1
2
PD•QH=

www.qhpta.com(八):

(2011•嘉兴一模)把Rt△ABC和Rt△DEF按如图(1)摆放(点C与E重合),点B、C(E)、F在同一条直线上.已知:∠ACB=∠EDF=90°,∠DEF=45°,AC=8cm,BC=6cm,EF=10cm.如图(2),△DEF从图(1)的位置出发,以1cm/s的速度沿CB向△ABC匀速移动,在△DEF移动的同时,点P从△ABC的顶点A出发,以2cm/s的速度沿AB向点B匀速移动;当点P移动到点B时,点P停止移动,△DEF也随之停止移动.DE与AC交于点Q,连接PQ,设移动时间为t(s).
(1)用含t的代数式表示线段AP和AQ的长,并写出t的取值范围;
(2)连接PE,设四边形APEQ的面积为y(cm2),试探究y的最大值;
(3)当t为何值时,△APQ是等腰三角形.

(1)AP=2t
∵∠EDF=90°,∠DEF=45°,
∴∠CQE=45°=∠DEF,
∴CQ=CE=t,
∴AQ=8-t,
t的取值范围是:0≤t≤5;

(2)过点P作PG⊥x轴于G,可求得AB=10,SinB=

4
5
,PB=10-2t,EB=6-t,
∴PG=PBSinB=
4
5
(10-2t)
∴y=S△ABC-S△PBE-S△QCE=
1
2
×6×8−
1
2
(6−t)×
4
5
(10−2t)−
1
2
t2
=
13
10
t2+
44
5
t=−
13
10
(t−
44
13
)2+
968
65

∴当t=
44
13
(在0≤t≤5内),y有最大值,y最大值=
968
65
(cm2

(3)若AP=AQ,则有2t=8-t解得:t=
8
3
(s)
若AP=PQ,如图①:过点P作PH⊥AC,则AH=QH=
8−t
2
,PH∥BC
∴△APH∽△ABC,
AP
AH
AB
AC

2t
8−t
2
10
8

解得:t=
40
21
(s)
若AQ=PQ,如图②:过点Q作QI⊥AB,则AI=PI=
1
2
AP=t
∵∠AIQ=∠ACB=90°∠A=∠A,
∴△AQI∽△ABC
AI
AQ
AC
AB
t
8−t
8
10

解得:t=
32
9
(s)
综上所述,当t=
8
3
40
21

www.qhpta.com(九):

(2014•张家界)如图,在平面直角坐标系中,O为坐标原点,抛物线y=ax2+bx+c(a≠0)过O、B、C三点,B、C坐标分别为(10,0)和(
32
9
18
5
【www.qhpta.com】

(1)设直线BC的解析式为y=kx+b,
∵直线BC经过B、C,

0=10k+b
24
5
18
5
k+b

解得:
k=
3
4
b=
−15
2

∴直线BC的解析式为;y=
3
4
x-
15
2


(2)∵抛物线y=ax2+bx+c(a≠0)过O、B、C三点,B、C坐标分别为(10,0)和(
18
5
,-
24
5
),
c=0
0=102a+10b+c
24
5
=(
18
5
)2a+
1
3

(1)如图1,过Q点作QP⊥BD交DC于P,
∴∠PQB=90°.
∵∠MQN=90°,
∴∠NQP=∠MQB,
∵四边形ABCD是正方形,
∴CD=CB,∠BDC=∠DBC=45°.DO=BO
∴∠DPQ=45°,DQ=PQ.
∴∠DPQ=∠DBC,
∴△QPN∽△QBM,

NP
MB
PQ
QB

∵Q是OD的中点,且PQ⊥BD,
∴DO=2DQ,DP=
1
2
DC
∴BQ=3DQ.DN+NP=
1
2
BC,
∴BQ=3PQ,
NP
MB
1
3

∴NP=
1
3
BM.
∴DN+
1
3
BM=
1
2
BC.

(2)如图2,过Q点作QH⊥BD交BC于H,
∴∠BQH=∠DQH=90°,
∴∠BHQ=45°.
∵∠COB=45°,
∴QH∥OC.
∵Q是OB的中点,
∴BH=CH=
1
2
BC.
∵∠NQM=90°,
∴∠NQD=∠MQH,
∵∠QND+∠NQD=45°,∠MQH+∠QMH=45°
∴∠QND=∠QMH,
∴△QHM∽△QDN,
HM
ND
QH
DQ
QM
NQ
1
3

∴HM=
1
3
ND,
∵BM-HM=HB,
BM−
1
3
DN=
1
2
BC

答案为:BM−
1
3
DN=
1
2
BC


(3)∵MB:MC=3:1,设CM=x,
∴MB=3x,
∴CB=CD=4x,
∴PB=2x,
∴PM=x.
∵HM=
1
3
ND,
∴ND=3x,
∴CN=7x
∵四边形ABCD是正方形,
∴ED∥BC,
∴△NDE∽△NCM,△DEF∽△BMF,
ND
CN
DE
CM
NE
NM
DE
BM
EF
FM

3x
7x
DE
x
NE
NM
3
7

∴DE=
3
7
x

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