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【发现问题】
如图11,点PP在等边三角形ABCABC内,且APC=150\angle APC=150^{\circ},PA=3PA=3,PC=4PC=4,求PBPB的长.小明发现,以APAP为边作等边三角形APDAPD,连接BDBD,得到ABD\triangle ABD;由等边三角形的性质,可证ACP\triangle ACPABD,\triangle ABD,PC=BDPC=BD;由已知APC=150\angle APC=150^{\circ},可知PDB\angle PDB的大小,进而可求得PBPB的长.
(1)(1)请回答:在图11中,PDB=\angle PDB=______,PB=______.PB=\_\_\_\_\_\_.
【问题解决】
(2)(2)参考小明思考问题的方法,解决下面问题:
如图22,ABC\triangle ABC中,ACB=90\angle ACB=90^{\circ},AC=BCAC=BC,点PPABC\triangle ABC内,且PA=1PA=1,PB=17PB=\sqrt{17},PC=22PC=2\sqrt{2},求APC\angle APCACAC的长.
【灵活运用】
(3)(3)如图33,某公园中有一块四边形空地ABCDABCD,连接ACAC,BDBD.已知AB=BDAB=BD,ABD=90\angle ABD=90^{\circ},BC=62BC=6\sqrt{2}米,DC=9DC=9米,公园规划部计划在四边形ABCDABCD内种植郁金香以供游客观赏,并将ACAC修建成观赏栈道,为保证观赏效果,要使ACAC的长度尽可能大(AC(AC的宽度不计),求此时种植郁金香的面积.
知识点:三角形章节:第13章 三角形 / 13.1 三角形的概念

答案与解析

答案

(1)APD\left(1\right)\because \triangle APD为等边三角形,ADP=60\therefore \angle ADP=60^{\circ}AP=DP=3AP=DP=3
ACP\because \triangle ACPABD,APC=150\triangle ABD,\angle APC=150^{\circ}
ADB=150\therefore \angle ADB=150^{\circ}BD=PC=4BD=PC=4
PDB=ADBADP=90\therefore \angle PDB=\angle ADB-\angle ADP=90^{\circ}.
RtBDPRt\triangle BDP中,BP=DP2+BD2=5BP=\sqrt{DP^{2}+BD^{2}}=5.
故答案为:9090^{\circ}55.
(2)(2)如图,将CPA\triangle CPA绕点CC逆时针旋转9090^{\circ}得到CGB\triangle CGB,连接PGPG.

CPA\because \triangle CPACBG\triangle CBG
CP=CG=22\therefore CP=CG=2\sqrt{2}CPG=CGP=45\angle CPG=\angle CGP=45^{\circ}AP=BG=1AP=BG=1
PG=2CP=4\therefore PG=\sqrt{2}CP=4.
PA=1\because PA=1PB=17PB=\sqrt{17}
PG2+BG2=PB2\therefore PG^{2}+BG^{2}=PB^{2}
PBG\therefore \triangle PBG是直角三角形,且PGB=90\angle PGB=90^{\circ}.
CGP=45\because \angle CGP=45^{\circ}
CGB=45+90=135\therefore \angle CGB=45^{\circ}+90^{\circ}=135^{\circ}
APC=135\therefore \angle APC=135^{\circ}.
此时APC+CPG=180\angle APC+\angle CPG=180^{\circ},即AAPPGG三点共线.
AG=AP+PG=5\therefore AG=AP+PG=5
RtABGRt\triangle ABG中,AB=AG2+BG2=26AB=\sqrt{AG^{2}+BG^{2}}=\sqrt{26}
ACB\because \triangle ACB为等腰直角三角形,
AC=13\therefore AC=\sqrt{13}.
综上:APC=135\angle APC=135^{\circ}AC=13AC=\sqrt{13}
(3)(3)如图,过BBBPBCBP\bot BC,使BP=BCBP=BC.

ABP+PBD=CBD+PBD=90\because \angle ABP+\angle PBD=\angle CBD+\angle PBD=90^{\circ}
ABP=CBD\therefore \angle ABP=\angle CBD
AB=BD\because AB=BD
ABP\therefore \triangle ABPDBC(SAS)\triangle DBC\left(SAS\right)
AP=CD=9m\therefore AP=CD=9m
CP=BC2+BP2=12m\because CP=\sqrt{BC^{2}+BP^{2}}=12m
\thereforeAPC\triangle APC中,根据三边关系可得:
PCAPACPC+AP(三点共线时取等)PC-AP\leqslant AC\leqslant PC+AP(三点共线时取等),即3AC213\leqslant AC\leqslant 21
\therefore如图,当APCAPC三点共线时,AC=21mAC=21m最大.

BAP+CAP+ADB=90\because \angle BAP+\angle CAP+\angle ADB=90^{\circ}BAP=CDB\angle BAP=\angle CDB
CDB+CAP+ADB=90\therefore \angle CDB+\angle CAP+\angle ADB=90^{\circ}
ACD=90\therefore \angle ACD=90^{\circ}.
S四边形ABCD=SABC+SACD\therefore S_{四边形ABCD}=S_{\triangle ABC}+S_{\triangle ACD}
BBBHBCBH\bot BC于点HH,由BCP\triangle BCP是等腰直角三角形易得,BH=6mBH=6m
S四边形ABCD=SABC+SACD=12ACBH+12ACCD=157.5m2\therefore S_{四边形ABCD}=S_{\triangle ABC}+S_{\triangle ACD}=\frac{1}{2}AC\cdot BH+\frac{1}{2}AC\cdot CD=157.5m^{2}.
\thereforeACAC最大时,四边形ABCDABCD的面积为157.5m2157.5m^{2}.

解析

(1)APD\left(1\right)\because \triangle APD为等边三角形,ADP=60\therefore \angle ADP=60^{\circ}AP=DP=3AP=DP=3
ACP\because \triangle ACPABD,APC=150\triangle ABD,\angle APC=150^{\circ}
ADB=150\therefore \angle ADB=150^{\circ}BD=PC=4BD=PC=4
PDB=ADBADP=90\therefore \angle PDB=\angle ADB-\angle ADP=90^{\circ}.
RtBDPRt\triangle BDP中,BP=DP2+BD2=5BP=\sqrt{DP^{2}+BD^{2}}=5.
故答案为:9090^{\circ}55.
(2)(2)如图,将CPA\triangle CPA绕点CC逆时针旋转9090^{\circ}得到CGB\triangle CGB,连接PGPG.

CPA\because \triangle CPACBG\triangle CBG
CP=CG=22\therefore CP=CG=2\sqrt{2}CPG=CGP=45\angle CPG=\angle CGP=45^{\circ}AP=BG=1AP=BG=1
PG=2CP=4\therefore PG=\sqrt{2}CP=4.
PA=1\because PA=1PB=17PB=\sqrt{17}
PG2+BG2=PB2\therefore PG^{2}+BG^{2}=PB^{2}
PBG\therefore \triangle PBG是直角三角形,且PGB=90\angle PGB=90^{\circ}.
CGP=45\because \angle CGP=45^{\circ}
CGB=45+90=135\therefore \angle CGB=45^{\circ}+90^{\circ}=135^{\circ}
APC=135\therefore \angle APC=135^{\circ}.
此时APC+CPG=180\angle APC+\angle CPG=180^{\circ},即AAPPGG三点共线.
AG=AP+PG=5\therefore AG=AP+PG=5
RtABGRt\triangle ABG中,AB=AG2+BG2=26AB=\sqrt{AG^{2}+BG^{2}}=\sqrt{26}
ACB\because \triangle ACB为等腰直角三角形,
AC=13\therefore AC=\sqrt{13}.
综上:APC=135\angle APC=135^{\circ}AC=13AC=\sqrt{13}
(3)(3)如图,过BBBPBCBP\bot BC,使BP=BCBP=BC.

ABP+PBD=CBD+PBD=90\because \angle ABP+\angle PBD=\angle CBD+\angle PBD=90^{\circ}
ABP=CBD\therefore \angle ABP=\angle CBD
AB=BD\because AB=BD
ABP\therefore \triangle ABPDBC(SAS)\triangle DBC\left(SAS\right)
AP=CD=9m\therefore AP=CD=9m
CP=BC2+BP2=12m\because CP=\sqrt{BC^{2}+BP^{2}}=12m
\thereforeAPC\triangle APC中,根据三边关系可得:
PCAPACPC+AP(三点共线时取等)PC-AP\leqslant AC\leqslant PC+AP(三点共线时取等),即3AC213\leqslant AC\leqslant 21
\therefore如图,当APCAPC三点共线时,AC=21mAC=21m最大.

BAP+CAP+ADB=90\because \angle BAP+\angle CAP+\angle ADB=90^{\circ}BAP=CDB\angle BAP=\angle CDB
CDB+CAP+ADB=90\therefore \angle CDB+\angle CAP+\angle ADB=90^{\circ}
ACD=90\therefore \angle ACD=90^{\circ}.
S四边形ABCD=SABC+SACD\therefore S_{四边形ABCD}=S_{\triangle ABC}+S_{\triangle ACD}
BBBHBCBH\bot BC于点HH,由BCP\triangle BCP是等腰直角三角形易得,BH=6mBH=6m
S四边形ABCD=SABC+SACD=12ACBH+12ACCD=157.5m2\therefore S_{四边形ABCD}=S_{\triangle ABC}+S_{\triangle ACD}=\frac{1}{2}AC\cdot BH+\frac{1}{2}AC\cdot CD=157.5m^{2}.
\thereforeACAC最大时,四边形ABCDABCD的面积为157.5m2157.5m^{2}.

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