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阅读材料:如图(1)\left(1\right),在AOB\triangle AOB中,O=90\angle O=90^{\circ},OA=OBOA=OB,点PPABAB边上,PEOAPE\bot OA于点EE,PFOBPF\bot OB于点FF,则PE+PF=OA.(此结论不必证明,可直接应用)PE+PF=OA.(此结论不必证明,可直接应用)

(1)(1)【理解与应用】
如图(2),正方形ABCDABCD的边长为22,对角线ACACBDBD相交于点OO,点PPABAB边上,PEOAPE\bot OA于点EE,PFOBPF\bot OB于点FF,则PE+PF=PE+PF=______;
(2)(2)【类比与推理】
如图(3)\left(3\right),矩形ABCDABCD的对角线ACACBDBD相交于点OO,AB=4AB=4,AD=3AD=3,点PPABAB边上,PE,PEOBOBACAC于点E,PFE,PFOAOABDBD于点FF,求PE+PFPE+PF的值;
(3)(3)【拓展与延伸】
四边形ABCDABCD是半径为44的圆内接四边形,对角线ACACBDBD相交于点OO,AB=AD=CDAB=AD=CD,点PP在弦BCBC,PE,PEACACBDBD于点E,PFE,PFBDBDACAC于点FF,当ABC=90\angle ABC=90^{\circ}时,试判断PE+PFPE+PF的值是否为定值,若是请求出该定值并求出四边形PEOFPEOF面积的最大值;若不是定值,请说明理由.
知识点:矩形的性质、正方形的性质、等边三角形的判定与性质、弦切角定理、圆的综合题、相似三角形的判定与性质章节:第22章 圆(下) / 一 直线和圆 / 22.2 圆的切线

答案与解析

答案

(1)\left(1\right)\because四边形ABCDABCD是正方形,
OA=OB=OC=OD\therefore OA=OB=OC=ODABC=AOB=90\angle ABC=\angle AOB=90^{\circ}.
AB=BC=2\because AB=BC=2
AC=22\therefore AC=2\sqrt{2}.
OA=2\therefore OA=\sqrt{2}.
OA=OB\because OA=OBAOB=90\angle AOB=90^{\circ}PEOAPE\bot OAPFOBPF\bot OB
PE+PF=OA=2\therefore PE+PF=OA=\sqrt{2}.
(2)(2)\because四边形ABCDABCD是矩形,
OA=OB=OC=OD\therefore OA=OB=OC=ODDAB=90\angle DAB=90^{\circ}
AB=4\because AB=4AD=3AD=3
BD=5\therefore BD=5
OA=OB=OC=OD=52\therefore OA=OB=OC=OD=\frac{5}{2}
PE\because PEOB,PFOB,PFAOAO
AEP\therefore \triangle AEPAOB,BFP\triangle AOB,\triangle BFPBOA\triangle BOA
EPOB=APABFPOA=BPAB\therefore \frac{EP}{OB}=\frac{AP}{AB},\frac{FP}{OA}=\frac{BP}{AB}
EPOB+FPOA=APAB+BPAB=1\therefore \frac{EP}{OB}+\frac{FP}{OA}=\frac{AP}{AB}+\frac{BP}{AB}=1
EP52+FP52=1\therefore \frac{EP}{\frac{5}{2}}+\frac{FP}{\frac{5}{2}}=1.
EP+FP=52\therefore EP+FP=\frac{5}{2}.
(3)(3)ABC=90\angle ABC=90^{\circ}时,PE+PFPE+PF是定值.理由如下:
连接OAOAOBOBOCOCODOD,如图33

\because四边形ABCDABCD是的圆内接四边形,
ADC=18090=90\therefore \angle ADC=180^{\circ}-90^{\circ}=90^{\circ}OA=OB=OC=ODOA=OB=OC=OD
AC\because ACBDBD是四边形ABCDABCD的对角线,对角线ACACBDBD相交于点OO
DAB=ADC=DCB=ABC=90\therefore \angle DAB=\angle ADC=\angle DCB=\angle ABC=90^{\circ}
\therefore四边形ABCDABCD是矩形,
OA=OB=OC=OD=12AC=12BD=4\therefore OA=OB=OC=OD=\frac{1}{2}AC=\frac{1}{2}BD=4
AB=AD=CD\because AB=AD=CD
\therefore四边形ABCDABCD是正方形,
PE\because PEAC,PFAC,PFBDBD
BOC=90\therefore \angle BOC=90^{\circ}PEOBPE\bot OBPFOCPF\bot OC
PE+PF=OC=4\therefore PE+PF=OC=4
PE=xPE=x,则PF=4xPF=4-x
\because四边形PEOFPEOF是矩形,
\therefore四边形PEOFPEOF面积=PEPF=x(4x)=x2+4x=(x2)2+4=PE\cdot PF=x\left(4-x\right)=-x^{2}+4x=-\left(x-2\right)^{2}+4
1<0\because -1 \lt 0
\thereforePE=x=2PE=x=2时,四边形PEOFPEOF面积有最大值,最大值为44.

解析

(1)\left(1\right)\because四边形ABCDABCD是正方形,
OA=OB=OC=OD\therefore OA=OB=OC=ODABC=AOB=90\angle ABC=\angle AOB=90^{\circ}.
AB=BC=2\because AB=BC=2
AC=22\therefore AC=2\sqrt{2}.
OA=2\therefore OA=\sqrt{2}.
OA=OB\because OA=OBAOB=90\angle AOB=90^{\circ}PEOAPE\bot OAPFOBPF\bot OB
PE+PF=OA=2\therefore PE+PF=OA=\sqrt{2}.
(2)(2)\because四边形ABCDABCD是矩形,
OA=OB=OC=OD\therefore OA=OB=OC=ODDAB=90\angle DAB=90^{\circ}
AB=4\because AB=4AD=3AD=3
BD=5\therefore BD=5
OA=OB=OC=OD=52\therefore OA=OB=OC=OD=\frac{5}{2}
PE\because PEOB,PFOB,PFAOAO
AEP\therefore \triangle AEPAOB,BFP\triangle AOB,\triangle BFPBOA\triangle BOA
EPOB=APABFPOA=BPAB\therefore \frac{EP}{OB}=\frac{AP}{AB},\frac{FP}{OA}=\frac{BP}{AB}
EPOB+FPOA=APAB+BPAB=1\therefore \frac{EP}{OB}+\frac{FP}{OA}=\frac{AP}{AB}+\frac{BP}{AB}=1
EP52+FP52=1\therefore \frac{EP}{\frac{5}{2}}+\frac{FP}{\frac{5}{2}}=1.
EP+FP=52\therefore EP+FP=\frac{5}{2}.
(3)(3)ABC=90\angle ABC=90^{\circ}时,PE+PFPE+PF是定值.理由如下:
连接OAOAOBOBOCOCODOD,如图33

\because四边形ABCDABCD是的圆内接四边形,
ADC=18090=90\therefore \angle ADC=180^{\circ}-90^{\circ}=90^{\circ}OA=OB=OC=ODOA=OB=OC=OD
AC\because ACBDBD是四边形ABCDABCD的对角线,对角线ACACBDBD相交于点OO
DAB=ADC=DCB=ABC=90\therefore \angle DAB=\angle ADC=\angle DCB=\angle ABC=90^{\circ}
\therefore四边形ABCDABCD是矩形,
OA=OB=OC=OD=12AC=12BD=4\therefore OA=OB=OC=OD=\frac{1}{2}AC=\frac{1}{2}BD=4
AB=AD=CD\because AB=AD=CD
\therefore四边形ABCDABCD是正方形,
PE\because PEAC,PFAC,PFBDBD
BOC=90\therefore \angle BOC=90^{\circ}PEOBPE\bot OBPFOCPF\bot OC
PE+PF=OC=4\therefore PE+PF=OC=4
PE=xPE=x,则PF=4xPF=4-x
\because四边形PEOFPEOF是矩形,
\therefore四边形PEOFPEOF面积=PEPF=x(4x)=x2+4x=(x2)2+4=PE\cdot PF=x\left(4-x\right)=-x^{2}+4x=-\left(x-2\right)^{2}+4
1<0\because -1 \lt 0
\thereforePE=x=2PE=x=2时,四边形PEOFPEOF面积有最大值,最大值为44.

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