青島出版社八年級(jí)寒假生活指導(dǎo)答案

字號(hào):

一、選擇題
    題號(hào) 1 2 3 4 5 6 7 8 9 10
    答案 C D B D A C D D D B
    二、填空題
    11、 12cm 12、 140° 和 50° 13、 540 ° 14、 45°
    15、 8(5.0 )或 (-5.0 ) 或 (8.0 ) 或 ( 0,5 )或(0,6) 16、 108°
    17證明:∵AB=AC,
    ∴∠B=∠C,
    在△ABD與△ACE中,
    ∵ ,
    ∴△ABD≌△ACE(SAS),
    ∴AD=AE.
    18:解:由題意知AB∥DE,
    ∴ ∠B=∠D
    在△BCA和△DCE中
    ∠B=∠D
    BC=DC
    ∠BCA=∠DCE
    ∴△BCA=△DCE(AAS)
    ∴ AB=DE
    19:過(guò)D點(diǎn)作DF//BE
    ∴∠ABC=∠DFC ∠E =∠ODF1
    ∵AB=AC
    ∴∠ABC=∠C
    ∴∠DFC=∠C
    ∴DF=DC
    ∵BE=DC
    ∴DF=BE-4
    在△EBO和△DFO中
    ∠E=∠ODF
    ∠BOE=∠D0F
    BE=DF
    △EBO≌△DFO(AAS)
    OE=OD6
    20:證明:∵△ABC和△ADE都是等腰直角三角形
    ∴AD=AE AB=AC2
    又∵∠EAC=90°+∠CAD, ∠DAB=90°+∠CAD
    ∴∠DAB=∠EAC4
    在△ADB和△AEC中
    AD=AE
    ∠DAB=∠EAC
    AB=AC
    ∴△ADB≌△AEC(SAS) 7
    ∴BD=CE8
    21證明:(1)∵AB=AC,D是BC的中點(diǎn),
    ∴∠BAE=∠EAC,
    在△ABE和△ACE中, ,
    ∴△ABE≌△ACE(SAS),
    ∴BE=CE;-3
    (2)∵∠BAC=45°,BF⊥AF,
    ∴△ABF為等腰直角三角形,
    ∴AF=BF,
    ∵AB=AC,點(diǎn)D是BC的中點(diǎn),
    ∴AD⊥BC,
    ∴∠EAF+∠C=90°,
    ∵BF⊥AC,
    ∴∠CBF+∠C=90°,
    ∴∠EAF=∠CBF,
    在△AEF和△BCF中,
    ∴△AEF≌△BCF(ASA).-8
    22:證明:∵AB∥CD
    ∴∠BAC=∠DCA
    在△BAC和△DCA中,
    AB=CD
    ∠BAC=∠DCA
    AC=CA
    △BAC≌△DCA(SAS)
    ∴∠DAC=∠BCA
    ∴ AD//BC4
    OE=OF
    由得∠E =∠F
    ∵O是AC的中點(diǎn)
    ∴OA=OC
    在△AOE和△COF中,
    ∠E =∠F
    ∠AOE=∠COF
    OA=OC
    △AOE≌△COF(AAS)
    ∴OE=OF-8
    23:(1)∵AB∥CD∠BED是△ABE的一個(gè)外角,
    ∴∠BED=∠ABE+∠BAD=15°+40°=55°。-3
    (2)如圖所示,EF即是△BED中BD邊上的高
    5
    (3)∵AD為△ABC的中線,BE為△ABD的中線,
    ∴S△ABD= S△ABC,S△BDE= S△ABD,
    ∴S△BDE= S△ABC,
    ∵△ABC的面積為40,BD=5,
    ∴S△BDE= BD•EF= ×5•EF= ×40,
    解得:EF=4-10
    25證明:(1)∵BD⊥直線m,CE⊥直線m
    ∴∠BDA=∠CEA=90°
    ∵∠BAD+∠ABD=90°
    ∴∠CAE=∠ABD1
    又AB=AC
    ∴△ADB≌△CEA2
    ∴AE=BD,AD=CE
    ∴DE=AE+AD= BD+CE 3
    (2)∵∠BDA =∠BAC= ,
    ∴∠DBA+∠BAD=∠BAD +∠CAE=180°—
    ∴∠DBA=∠CAE4
    ∵∠BDA=∠AEC= ,AB=AC
    ∴△ADB≌△CEA5
    ∴AE=BD,AD=CE
    ∴DE=AE+AD=BD+CE:7
    (3)由(2)知,△ADB≌△CEA,
    BD=AE,∠DBA =∠CAE
    ∵△ABF和△ACF均為等邊三角形
    ∴∠ABF=∠CAF=60°
    ∴∠DBA+∠ABF=∠CAE+∠CAF
    ∴∠DBF=∠FAE9
    ∵BF=AF
    ∴△DBF≌△EAF10
    ∴DF=EF,∠BFD=∠AFE
    ∴∠DFE=∠DFA+∠AFE=∠DFA+∠BFD=60°
    ∴△DEF為等邊三角形.12
    6答:
    解:如圖,連接OB、OC,
    ∵∠BAC=54°,AO為∠BAC的平線,
    ∴∠BAO=∠BAC=×54°=27°,
    又∵AB=AC,
    ∴∠ABC=(180°﹣∠BAC)=(180°﹣54°)=63°,
    ∵DO是AB的垂直平線,
    ∴OA=OB,
    ∴∠ABO=∠BAO=27°,
    ∴∠OBC=∠ABC﹣∠ABO=63°﹣27°=36°,
    ∵DO是AB的垂直平線,AO為∠BAC的平線,
    ∴點(diǎn)O是△ABC的外心,
    ∴OB=OC,
    ∴∠OCB=∠OBC=36°,
    ∵將∠C沿EF(E在BC上,F(xiàn)在AC上)折疊,點(diǎn)C與點(diǎn)O恰好重合,
    ∴OE=CE,
    ∴∠COE=∠OCB=36°,
    在△OCE中,∠OEC=180°﹣∠COE﹣∠OCB=180°﹣36°﹣36°=108°.
    故答案為:108.
    9解答:解:作B點(diǎn)關(guān)于y軸對(duì)稱點(diǎn)B′點(diǎn),連接AB′,交y軸于點(diǎn)C′,
    此時(shí)△ABC的周長(zhǎng)最小,
    ∵點(diǎn)A、B的坐標(biāo)別為(1,4)和(3,0),
    ∴B′點(diǎn)坐標(biāo)為:(﹣3,0),AE=4,
    則BE=4,即BE=AE,
    ∵C′O∥AE,
    ∴B′O=C′O=3,
    ∴點(diǎn)C′的坐標(biāo)是(0,3),此時(shí)△ABC的周長(zhǎng)最小.
    故選:D.
    10:解答: 解:設(shè)∠A=x,
    ∵AP1=P1P2=P2P3==P13P14=P14A,
    ∴∠A=∠AP2P1=∠AP13P14=x,
    ∴∠P2P1P3=∠P13P14P12=2x,
    ∴∠P2P3P4=∠P13P12P10=3x,
    ∠P7P6P8=∠P8P9P7=7x,
    ∴∠AP7P8=7x,∠AP8P7=7x,
    在△AP7P8中,∠A+∠AP7P8+∠AP8P7=180°,
    即x+7x+7x=180°,
    解得x=12°,
    即∠A=12°.
    故答案為:12°.