初二数学，动点问题
\u521d\u4e8c\u6570\u5b66\u52a8\u70b9\u95ee\u9898\u6c42\u4e09\u89d2\u5f62AQP\u7684\u9762\u79ef\uff0c\u5b9e\u9645\u4e0aAQP\u7684\u9762\u79ef\u7b49\u4e8e\u56db\u8fb9\u5f62ABCD\u7684\u9762\u79ef\u51cf\u53bb\u4e09\u4e2a\u5c0f\u4e09\u89d2\u5f62\u7684\u9762\u79ef
1.p\u70b9\u7684\u8fd0\u52a8\u901f\u5ea6\u662f1\u4e2a\u5355\u4f4d/\u79d2\uff0c\u51fa\u53d1x\u79d2\uff0c\u6240\u4ee5BP=x
S\u25b3AQP=S\u957f\u65b9\u5f62ABCD-S\u25b3ABP-S\u25b3ADQ-S\u25b3CPQ
=AB*BC-1/2*AB*BP-1/2*AD*DQ-1/2*CP*CQ
=8-x-2-(4-x)/2
=4-x/2
BC=4\uff0cP\u6cbfBC\u8fd0\u52a8\uff0cBC/1=4\u79d2
\u6240\u4ee5x\u7684\u53d6\u503c\u8303\u56f4\u4e3a0\u2264x\u22644
2.\u4e09\u89d2\u5f62AQP\u662f\u7b49\u8170\u4e09\u89d2\u5f62\uff0c\u90a3\u4e48\u6709\u4e09\u79cd\u60c5\u51b5\uff0cAQ=QP\uff0cAP=PQ\uff0cAP=AQ
\u7b2c\u4e00\u79cd\u60c5\u51b5\uff0cAQ=QP\uff0c\u4ece\u56fe\u5f62\u4e0a\u4e0d\u96be\u770b\u51fa\uff0c\u8fd9\u65f6\u5019P\u548cB\u6b63\u597d\u91cd\u5408\uff0c\u5373x=0
\u7b2c\u4e8c\u79cd\u60c5\u51b5\uff0cAP=PQ\uff0cAP=\u6839\u53f7\uff08AB²+BP²\uff09\uff0cPQ=\u6839\u53f7\uff08CP²+CQ²\uff09\uff0cBP=x\uff0cCP=4-x\uff0cCQ=1
\u89e3\u5f97x=13/8
\u7b2c\u4e09\u79cd\u60c5\u51b5\uff0cAP=AQ\uff0cAQ=\u6839\u53f7\uff08AD²+DQ²\uff09=\u6839\u53f717
AP=\u6839\u53f7\uff08AB²+BP²\uff09=\u6839\u53f717
x=\u6839\u53f713
1\uff0c\u5206\u6790\u77e5\u6b64\u65f62t+t=5 t=5/3 (\u8fc7C\u70b9\u505aAB\u5782\u7ebfCF\uff0c\u6c42\u51faAF=5\u5373\u53ef\uff09
2,\u6c42\u51fa\u68af\u5f62\u9ad8\u4e3a3\u221a3\uff0c\u9762\u79ef\u4e3a15\u221a3
\u5f53Q\u5728DC\u4e0a\uff0ct\u22642.\u9762\u79ef\u4e3a(t+8-2t)/2\u00d73\u221a3=15\u221a3/2 \uff0c\u5f97t=3\u4e0d\u7b26\u5408\uff0c\u6545\u6392\u9664\u8be5\u53ef\u80fd\uff1b
\u5f53Q\u5728AD\u4e0a,2<t\u22644,\u6211\u4eec\u8ba1\u7b97\u4f59\u4e0b\u9762\u79ef\u7b49\u4e8e\u68af\u5f62\u4e00\u534a\u5373\u53ef\uff0c\u4f59\u4e0b\u4e24\u4e2a\u4e09\u89d2\u5f62DCQ\u548cAPQ\uff1b\u6b64\u65f6QD=t-2\uff0c\u90a3\u4e48QD/AD=(t-2)/6 ,\u4ee5\u6b64\u6c42\u51fa\u4e09\u89d2\u5f62DQC\u9ad8\u4e3a\u221a3(t-2)/2
\u9762\u79ef\u4e3a\u221a3(t-2)/2;\u540c\u7406\u6c42\u51fa\u4e09\u89d2\u5f62AQP\u9762\u79ef\u4e3a\u221a3t(8-t)/2;
\u4e24\u8005\u76f8\u52a0\u4e3a\u221a3/2(t-2+8t-t2)=15\u221a3/2 \u89e3\u51fat\u820d\u53bb\u4e0d\u7b26\u5408\u89e3\u5f97t=(9-\u221a13)/2
（2）解：∵△ABP≌△DCP∴BP=CP=1/2BC=5cm∴PC=10-2t=5∴t=5/2s
（3）解：【注意】题目说△ABP与△PQC全等，那么就有两种对应关系，要分类讨论。①△ABP≌△QCP∴CQ=AB=6cm BP=CP=5cm∴t=5/2s此时CQ=vt=65/2v=6∴v=12/5cm/s②△ABP≌△PCQ∴AB=CP=6cm=10-2t BP=CQ=2t∴t=2∴CQ=4cm∴v=2cm/s综上所述：v=12/5cm/s或v=2cm/s
望采纳！！！
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绛旓細鈭燘AC=90掳, 鎵浠BCD鏄煩褰傚綋AQCP鏄煩褰紝 褰揂QCP鏄彵褰㈡椂锛宷p骞冲垎鍨傜洿浜嶢C锛岃PQ 浜C浜嶮锛岃BQ= x锛屽嬀鑲″畾鐞 AB^2 + x^2 = AQ^2 AQ = QC = BC - BQ, => AB^2 + x^2 = (BC - BQ)^2,甯﹀叆鏁板 6^2 + x^2 = (8 - x)^2. 鎵浠 x = 7 / 4 ...
绛旓細瑙ｏ細鐢遍鎰忓彲鐭ワ細OA=1锛孫C=3锛孫B=鈭2 鈭磘an鈭燘AC =鈭3 鈭粹垹BAC =60掳 鍚岀悊鈭燗CB=30掳 鑻モ柍ABC涓庘柍ABP鐩镐技 鏈夊洓绉嶆儏鍐 锛1锛塒鍦–B鐨勫欢闀跨嚎涓婏紝BP/AB=1/鈭3,姝ゆ椂P鐐瑰潗鏍囦负锛1锛4/3鈭3锛夛紙2锛塒鍦–B鐨勫欢闀跨嚎涓婏紝BP/AB=鈭3,姝ゆ椂P鐐瑰潗鏍囦负锛3锛2鈭3锛夛紙3锛塒鍦–B涓婏紝BP/AB=1...
