初三数学几何题目 一道初三数学几何题
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\u8fd9\u9898\u65e0\u975e\u662f\u601d\u8def\u95ee\u9898\uff0c
\u505a\u8f85\u52a9\u7ebf\u8fde\u63a5BD\u4e0eAC\u76f8\u4ea4\u4e8eE\u3002
\u7531\u5bf9\u89d2\u7ebf\u53ef\u77e5AC=2AE
\u4e5f\u5c31\u662f\u76f8\u5f53\u4e8e\u6c42AE\u7684\u6700\u5c0f\u503c\u3002
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\u586b\u7a7a\u9898\uff0c\u53c8\u4e0d\u9700\u8981\u8bc1\u660e\u3002
\u89e3\uff1a(1)\u2235\u25b3ABC\u662f\u8fb9\u957f\u4e3a6\u7684\u7b49\u8fb9\u4e09\u89d2\u5f62\uff0c
\u2234\u2220ACB=60\u00b0\uff0c
\u2235\u2220BQD=30\u00b0\uff0c
\u2234\u2220QCP=90\u00b0\uff0c
\u8bbeAP=x\uff0c\u5219PC=6\ufe63x\uff0cQB=x\uff0c
\u2234QC=QB+C=6+x\uff0c
\u2235\u5728Rt\u25b3QCP\u4e2d\uff0c\u2220BQD=30\u00b0\uff0c
\u2234PC=½QC\uff0c\u53736\ufe63x=½(6+x)\uff0c\u89e3\u5f97x=2\uff1b
(2)\u5f53\u70b9P\u3001Q\u8fd0\u52a8\u65f6\uff0c\u7ebf\u6bb5DE\u7684\u957f\u5ea6\u4e0d\u4f1a\u6539\u53d8\uff0e\u7406\u7531\u5982\u4e0b\uff1a
\u4f5cQF\u22a5AB\uff0c\u4ea4\u76f4\u7ebfAB\u7684\u5ef6\u957f\u7ebf\u4e8e\u70b9F\uff0c\u8fde\u63a5QE\uff0cPF\uff0c
\u53c8\u2235PE\u22a5AB\u4e8eE\uff0c
\u2234\u2220DFQ=\u2220AEP=90\u00b0\uff0c
\u2235\u70b9P\u3001Q\u505a\u5300\u901f\u8fd0\u52a8\u4e14\u901f\u5ea6\u76f8\u540c\uff0c
\u2234AP=BQ\uff0c
\u2235\u25b3ABC\u662f\u7b49\u8fb9\u4e09\u89d2\u5f62\uff0c
\u2234\u2220A=\u2220ABC=\u2220FBQ=60\u00b0\uff0c
\u2234\u5728\u25b3APE\u548c\u25b3BQF\u4e2d\uff0c
\u2235\u2220A=\u2220FBQ\u2220AEP=\u2220BFQ=90\u00b0\uff0c
\u2234\u2220APE=\u2220BQF\uff0c
\u2234\u2220A=\u2220FBQ
AP=BQ
\u2220AEP=\u2220BFQ
\u2234\u25b3APE\u224c\u25b3BQF\uff0c
\u2234AE=BF\uff0cPE=QF\u4e14PE\u2225QF\uff0c
\u2234\u56db\u8fb9\u5f62PEQF\u662f\u5e73\u884c\u56db\u8fb9\u5f62\uff0c
\u2234DE=½EF\uff0c
\u2235EB+AE=BE+BF=AB\uff0c
\u2234DE=½AB\uff0c
\u53c8\u2235\u7b49\u8fb9\u25b3ABC\u7684\u8fb9\u957f\u4e3a6\uff0c
\u2234DE=3\uff0c
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连接OP,与AB交于M
∵PA、PB是圆O的切线
∴PA
=PB
且
PO平分∠APB
∴PO⊥AB
∴AM
=
BM
∵AO
=
BO
∴OM
=
1/2
BC
=
1/2
∵∠PAO
=
∠AMP
=
90°
∠OPA为公共角
∴△PAO∽△PMA
∴PA²
=
PM*PO
即:3
=
(PO
-
1/2)*PO
∴PO
=
2
∴AO
=
√(PO²
-
AP²)
=
1
故:圆的半径为1
1、连接OC,证明OC⊥CD就可以了。(因为OA=OC,所以∠OAC=∠OCA,因为AC平分∠BAD,所以∠OCA=∠CAD,内错角相等,所以OC//AD,因为AD⊥CD,所以OC⊥CD)
2、从O向AD作垂线,垂直于E,则OEDC是矩形,且OC=DE=2,因为AD=3,则AE=1,因为OA=2,所以直角三角形OAE中,∠AOE=30度,则∠OAE=60度,因为AC平分∠BAD,所以∠BAC=30度。
绛旓細2銆佽В锛氳繛鎺P锛屼笌AB浜や簬M 鈭礟A銆丳B鏄渾O鐨勫垏绾 鈭碢A =PB 涓 PO骞冲垎鈭燗PB 鈭碢O鈯B 鈭碅M = BM 鈭礎O = BO 鈭碠M = 1/2 BC = 1/2 鈭碘垹PAO = 鈭燗MP = 90掳 鈭燨PA涓哄叕鍏辫 鈭粹柍PAO鈭解柍PMA 鈭碢A²= PM*PO 鍗筹細3 = 锛圥O - 1/2)*PO 鈭碢O = 2 鈭碅O = 鈭氾紙PO...
绛旓細璇存槑锛氫綘鐨棰樼洰鏈夎锛佲淎B锛孊C锛孋D锛屽垎鍒簬鍦哋浜や簬鐐笶锛孎锛孏鈥濆簲璇ヤ负鈥淎B锛孊C锛孋D锛屽垎鍒垏鍦哋浜庣偣E锛孎锛孏鈥濊瘉鏄庯細鍥犱负AB,BC鏄渾鐨勫垏绾 鎵浠モ垹EBO=鈭燨BF=1/2鈭燗BC(渚濇嵁锛氬垏绾块暱瀹氱悊锛氫粠鍦嗗涓鐐瑰紩鍦嗙殑涓ゆ潯鍒囩嚎锛屽畠浠殑鍒囩嚎闀跨浉绛夛紝涓斿渾蹇冧笌鍦嗗杩欑偣鐨勮繛绾垮钩鍒嗕袱鍒囩嚎涔嬮棿鐨勫す瑙掞級鍚岀悊锛氣垹...
绛旓細瑙g瓟锛氾紙1锛夆憼鐚滄兂BG=DE,涓斾簩鑰呮墍鍦ㄧ殑鐩寸嚎鐩镐簰鍨傜洿銆傗埖鍥涜竟褰BCD涓庡洓杈瑰舰CEFG閮芥槸姝f柟褰傗埓BC=DC,CG=CE,鈭燘CG=鈭燚CE=90掳 鈭粹柍BCG鈭解柍DCE 鏁匓G=CE,鈭燘GC=鈭燚EC 鍙堚垹BGC+鈭燙BG=90掳 鈭粹垹DEC+鈭燙BG=90掳 BG涓嶥E鎵鍦ㄧ洿绾胯BC鎵鍦ㄧ洿绾挎墍鎴紝褰㈡垚鐨勫悓鏃佸唴瑙掍簰涓轰綑瑙掞紝鍒欑洿绾緽G鈯E.鈶′换鐒...
绛旓細鈭燗BC = 180- 3*锛坸+y锛180 - 3x - 3*30 = 90 -3x = 3*锛30-x锛=3鈭燚BC銆傦紙2锛夊綋sinx = 0鏃讹紝 x =0搴︺180搴︼紝璇存槑ACD鍦ㄤ竴鏉$洿绾夸笂锛孌鐐逛笉鍦ㄤ笁瑙掑舰鍐呫傦紙鍗充笉婊¤冻棰樼洰鏉′欢銆傚疄闄呬笂锛屽嵆浣緿鍦ˋC涓偣锛屼袱瑙掍篃婊¤冻3鍊嶅叧绯伙細姝ゆ椂瑙扐=60搴︺傦級杩欐牱灏卞彲浠ュ緱鍑轰笂杩扮殑缁撹浜嗐
绛旓細DE涓嶢C浜ょ偣鏄疓鍚 鍒橤E=5 GE//AB GE/AB=EC/BC EC=25/3锛孊C=40/3 S闃=S涓夎褰BC-S涓夎褰EC =1/2*8*40/3-1/2*5*25/3 =205/6
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绛旓細杩欓亾棰樼粨璁烘槸浜旇竟褰BCDE鐨勯潰绉负1 鍥犱负鏈変釜鍏崇郴锛孲=(BD²/2)*sin鈭燙DE=2sin30掳=1 涓嬮潰鏉ヨ瘉鏄庝竴鑸儏鍐碉細濡傚浘1鎵绀猴紝AB=BC锛孋D=DE鐨勫嚫浜旇竟褰紝璁锯垹CDE=伪锛屸垹ABC=尾锛屛=180掳-尾锛孊D=a銆傚皢鍥1涓殑鈻矰CB缁旸鐐归嗘椂閽堟棆杞卞悗寰楀埌鈻矰C'B'鈭礐D=DE 鈭碈D涓嶥E閲嶅悎锛孍鐐瑰嵆涓篊'鐐癸紝...
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绛旓細璇佹槑锛1銆佺敱AC鈭D鍙煡锛屸垹BAC+鈭燗BD=180掳锛屽嵆鈭燩AC+鈭燩AB+鈭燩BA+鈭燩BD=180掳 寰楋紝鈭燩AC+鈭燩BD=180掳-锛堚垹PAB+鈭燩BA锛夊張鈭燩AB+鈭燩BA+鈭燗PB=180掳锛屽嵆鈭燗PB=180掳-锛堚垹PAB+鈭燩BA锛夆埓鈭燗PB=鈭燩AC+鈭燩BD 2銆佸浘锛2锛夊涓婅堪寰楃粨璁轰笉鎴愮珛 3銆佲垹PAC=鈭燗PB+鈭燩BD 4銆佷笌鍥撅紙3锛変腑鐨勫叧绯...
绛旓細锛1锛夎瘉鏄庯細鍥犱负AB=AC 鎵浠ヤ笁瑙掑舰ABC鏄瓑鑵颁笁瑙掑舰 鍥犱负AD鍨傜洿BC浜嶥 鎵浠D鏄瓑鑵颁笁瑙掑舰ABC鐨勫瀭绾匡紝涓嚎锛岃骞冲垎绾 鎵浠ヨADC=90搴 BD=CD 瑙払AD=瑙扖AD=1/2瑙払AC 鍥犱负AN鏄BAC鐨勫瑙扖AM鐨勫钩鍒嗙嚎 鎵浠ヨCAN=瑙扢AN=1/2瑙扖AM 鍥犱负瑙払AC+瑙扖AM=180搴 鎵浠ヨCAD+瑙扖AN=瑙扗AN=90搴 鍥犱负CE鍨傜洿...