初中数学几何题
\u521d\u4e2d\u6570\u5b66\u9898\u76ee\uff0c\u51e0\u4f55\u9898\u3010\u9898\u76ee\u3011
\u5df2\u77e5\u5728\u25b3ABC\u4e2d\uff0c\u2220CAB=2\u03b1\uff0c\u4e140\uff1c\u03b1\uff1c30\u00b0\uff0cAP\u5e73\u5206\u2220CAB\uff0c\u82e5\u2220ABC=60\u00b0-\u03b1\uff0c\u70b9P\u5728\u25b3ABC\u7684\u5185\u90e8\uff0c\u4e14\u4f7f\u2220CBP=30\u00b0\uff0c\u6c42\u2220APC\u7684\u5ea6\u6570\uff08\u7528\u542b\u03b1\u7684\u4ee3\u6570\u5f0f\u8868\u793a\uff09\u3002
\u3010\u89e3\u7b54\u3011
\u3010\u89e3\u6cd5\u4e00\u3011
\u89e3\uff1a
\u5ef6\u957fAC\u81f3M\uff0c\u4f7fAM=AB\uff0c\u8fde\u63a5PM\uff0cBM\uff08\u5982\u56fe1\uff09
\u2235AP\u5e73\u5206\u2220CAB\uff0c\u2220CAB=2\u03b1
\u2234\u22201=\u22202= \u03b1
\u5728\u25b3AMP\u548c\u25b3ABP\u4e2d\uff1a
\u2235AM=AB\uff0c\u22201 =\u22202\uff0cAP=AP
\u2234\u25b3AMP\u224c\u25b3ABP
\u2234PM=PB\uff0c\u22203 =\u22204
\u2235\u2220ABC=60\u00b0\uff0d\u03b1\uff0c\u2220CBP=30\u00b0
\u2234\u22204=(60\u00b0\uff0d\u03b1)\uff0d30\u00b0=30\u00b0\uff0d\u03b1
\u2234\u22203 =\u22204 =30\u00b0\uff0d\u03b1
\u2235\u25b3AMB\u4e2d\uff0cAM=AB
\u2234\u2220AMB=\u2220ABM=(180\u00b0\uff0d\u2220MAB)\u00f72 =(180\u00b0\uff0d2\u03b1)\u00f72 =90\u00b0\uff0d\u03b1
\u2234\u22205=\u2220AMB\uff0d\u22203= (90\u00b0\uff0d\u03b1)\uff0d(30\u00b0\uff0d\u03b1)=60\u00b0
\u2234\u25b3PMB\u4e3a\u7b49\u8fb9\u25b3
\u2235\u22206=\u2220ABM\uff0d\u2220ABC = (90\u00b0\uff0d\u03b1)\uff0d(60\u00b0\uff0d\u03b1)=30\u00b0
\u2234\u22206=\u2220CBP
\u2234BC\u5e73\u5206\u2220PBM
\u2234BC\u5782\u76f4\u5e73\u5206PM
\u2234CP=CM
\u2234\u22207 =\u22203 = 30\u00b0\uff0d\u03b1
\u2234\u2220ACP=\u22207\uff0b\u22203=(30\u00b0\uff0d\u03b1)\uff0b(30\u00b0\uff0d\u03b1)=60\u00b0\uff0d2\u03b1
\u2234\u25b3ACP\u4e2d\uff0c\u2220APC=180\u00b0\uff0d\u22201\uff0d\u2220ACP
=180\u00b0\uff0d\u03b1\uff0d(60\u00b0\uff0d2\u03b1)
=120\u00b0\uff0b\u03b1
\u3010\u89e3\u6cd5\u4e8c\u3011
\u89e3\uff1a
\u5728AB\u4e0a\u622a\u53d6AM\uff0c\u4f7fAM=AC\uff0c\u8fde\u63a5PM\uff0c\u5ef6\u957fAP\u4ea4BC\u4e8eN\uff0c\u8fde\u63a5MN\uff08\u5982\u56fe2\uff09
\u2235AP\u5e73\u5206\u2220CAB\uff0c\u2220CAB=2\u03b1
\u2234\u22201=\u22202=\u03b1
\u5728\u25b3ACN\u548c\u25b3AMN\u4e2d\uff1a
\u2235AC=AM\uff0c\u22201 =\u22202\uff0c AN=AN
\u2234\u25b3ACN\u224c\u25b3AMN
\u2234\u22203 =\u22204
\u2235\u2220ABC=60\u00b0\uff0d\u03b1
\u2234\u22203=\u22202\uff0b\u2220NBA=\u03b1\uff0b(60\u00b0\uff0d\u03b1) =60\u00b0
\u2234\u22203 =\u22204 =60\u00b0
\u2234\u22205=180\u00b0\uff0d\u22203\uff0d\u22204=180\u00b0\uff0d60\u00b0\uff0d60\u00b0=60\u00b0
\u2234\u22204 =\u22205
\u2234NM\u5e73\u5206\u2220PNB
\u2235\u2220CBP=30\u00b0
\u2234\u22206=\u22203\uff0d\u2220NBP=60\u00b0\uff0d30\u00b0=30\u00b0
\u2234\u22206=\u2220NBP
\u2234NP=NB
\u2234NM\u5782\u76f4\u5e73\u5206PB
\u2234MP=MB
\u2234\u22207 =\u22208
\u2234\u22206\uff0b\u22207 =\u2220NBP\uff0b\u22208
\u5373\u2220NPM=\u2220NBM =60\u00b0\uff0d\u03b1
\u2234\u2220APM=180\u00b0\uff0d\u2220NPM =180\u00b0\uff0d(60\u00b0\uff0d\u03b1)=120\u00b0\uff0b\u03b1
\u5728\u25b3ACP\u548c\u25b3AMP\u4e2d\uff1a
\u2235AC=AM\uff0c \u22201 =\u22202\uff0c AP=AP
\u2234\u25b3ACP\u224c\u25b3AMP
\u2234\u2220APC=\u2220APM
\u2234\u2220APC=120\u00b0\uff0b\u03b1
\u8fd9\u91cc\u6709\u51e0\u4f55\u7684\u89e3\u9898\u65b9\u6cd5\u4e5f\u8bb8\u6709\u7528
解:分为两种情况求解:(1)AB=AD+BE+DE:
此时,容易证明△AEC∽△BCD
AE/BC=AC/BD
即:AC*BC=AC^2=(AB-4)*(AB-3)=AB^2-7AB+12
AB^2=2*AC^2
解得AB=12,DE=5
S△ABC=36
(2)AB=AE+BD-DE
此时,容易证明△ADC∽△BCE
AD/BC=AC/BE
AC*BC=AC^2=AD*BE=12,
AC=2根号3
S△ABC=6
AD=3 BE=4 DE=7 AB=14 AC=BC=7根号2
面积ABC=14
旋转ECB使AC与BC重合得三角形FCA 因为角FCD=角DCE CB=CF DC=CD所以CDF全等DCB 所以FD=DE=FA+AD=3+4=7 所以AE=FE=7+7=14 而AC=AB 所以由勾股定理得AC=BC=根号98 所以ABC的面积=49
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绛旓細闂涓锛鍒濅腑鏁板鍑犱綍棰 璁锯垹ABO=X 鈭礎BCD 鈭粹垹ABC=鈭燘CD=40o 鈭礎B=AO 鈭粹垹O=鈭燗BO=X 鈭燙AB=2X 鈭礐B=AB 鈭粹垹ACB=CAB=2X 鈭2X+40+x+x=180 鈭磝=35o 鈭粹垹COD=35o 闂浜岋細鍒濅腑鏁板棰樼洰锛屽嚑浣曢 銆愰鐩戝凡鐭ュ湪鈻矨BC涓紝鈭燙AB=2伪锛屼笖0锛溛憋紲30掳锛孉P骞冲垎鈭燙AB锛岃嫢鈭燗BC=60...
绛旓細1锛氳彵褰㈢殑涓ゆ潯瀵硅绾块暱鍒嗗埆鏄1cm鍜2cm,鍒欒繖涓彵褰㈢殑闈㈢Н鏄紙1cm²锛夈2锛氬湪姝f柟褰BCD鎵鍦ㄥ钩闈㈠唴瀛樺湪锛堜腑蹇冩垨閲嶅績锛夛紝浣垮畠涓庡悇瀹氱偣鏋勬垚鐨勬墍鏈変笁瑙掑舰閮戒负绛夎叞涓夎褰3锛氬湪骞宠鍥涜竟褰BCD涓紝DC=2AD锛孧鏄疍C鐨勪腑鐐癸紝鍒欒AMB绛変簬锛90掳锛夈傘愬垎鏋愩戣繃M浣淢N鈭D锛屽垯MADN鍜孧NBC鏄彵褰侫M鍜孊...
绛旓細瑙:鈭燘AC=180掳-鈭燗BC-鈭燗CB=50掳.杩炴帴AI,BI,CI,AE.鐐笽涓衡娍ABC鐨勫唴蹇,鍒:鈭燙AI=(1/2)鈭燙AB=25掳,鈭燗CI=(1/2)鈭燗CB=30掳.鈭粹垹AIC=180掳-(鈭燙AI+鈭燗CI)=180掳-(25掳+30掳)=125掳;鈭碘垹ADE=鈭燗BC=70掳;DE=AD.鈭粹垹DEA=鈭燚AE=(180掳-鈭燗DE)/2=(180掳-70掳)/2=55掳.鍒...
绛旓細浠C鐨勪腑鐐癸紙璁句负D锛変负鍦嗗績锛屼互AC/2涓哄崐寰勪綔鍦嗐傦紙濡傚浘锛夊洜PA^2+PC^2=AC^2, 寰楄APC=90搴︺侾蹇呭湪鍦嗕笂銆傚綋B銆丳銆丏涓夌偣涓绾挎椂锛孭B鏈夋渶灏忓 锛圖B=PB+PD,PD=AC/2涓瀹氾紝 BD涓虹嚎娈垫椂鏈鐭紝姝ゆ椂PB鏈夋渶灏忓硷級姝ゆ椂锛欳D=鈭3, BC=3 瑙扗CB=90搴 寰楋細瑙扗BC涓30搴︼紝瑙払DC涓60搴 鎵浠...
绛旓細(1)銆佲埖ABCD鏄钩琛屽洓杈瑰舰銆傗埓鈭燘=鈭燙DM,AB=CD,BC=AD,鈭礏N=NC.AM=DM,鈭碆N=DM,鍒欌娍ABN鈮屸娍CDM锛(2)銆佺敱AM=MD,BN=NC鍙煡MD=NC,MD鈭C,鈭碝NCD鏄钩琛屽洓杈瑰舰銆傝岀敱AM=NC,AM鈭C鍙煡ANCM涔熸槸骞宠鍥涜竟褰紝鈭碅N鈭C,鈭碘垹AND=90掳,鈭粹垹MOD=90掳鍒橫NCD杩樻槸鑿卞舰銆侻N=NC銆傗垹1=鈭燚NC=鈭2...
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绛旓細5.鈭礏D涓烘鏂瑰舰ABCD鐨勫瑙掔嚎锛屸埓AF=CF,鈻矨FD鈮屸柍CFD锛圫AS锛夆垹DAF=鈭燚CF銆傛槗璇佲柍ABE鈮屸柍DCE锛圫AS)鈭粹垹ABE=鈭燚CF=鈭燚AF,鈭.鈭燛AF+鈭燗EB=鈭燗BE+鈭燗EB=90掳鍗矨F鈯E 杩欎簺閮芥槸鍒濅腑棰
绛旓細鍙堝洜涓築E骞冲垎瑙掆垹ABC锛孍C骞冲垎鈭燗CB 鎵浠モ垹EBC=27.5锛屸垹ECB=37.5 鎵浠モ垹BEC=115 锛2锛夊洜涓衡垹ABC=55锛屸垹ACB=75 鎵浠モ垹CBF=62.5锛屸垹BCF=52.5 鎵浠モ垹BFC=65 锛3锛夊洜涓衡垹EBC=37.5 锛屸垹ACB=75 鎵浠モ垹BCF=鈭燗CA1=52.5 鎵浠モ垹A1CB=155 鎵浠モ垹A1=25 鎵浠モ垹A=2鈭燗1 锛4锛夆垹A=2A1=4A...