初二数学动点问题，如图，已知三角形ABC中，∠B＝90° 初二数学题 动点问题 如图Rt三角形ABC中，角ACB=90...
\u5982\u56fe\uff0c\u5df2\u77e5\u25b3ABC\u4e2d\uff0c\u2220B=90\u00b0\uff0cAB=8cm\uff0cBC=6cm\uff0cP\u3001Q\u5206\u522b\u4e3aAB\u3001BC\u8fb9\u4e0a\u7684\u52a8\u70b9\uff0c\u70b9P\u4ece\u70b9A\u5f00\u59cb\u6cbfA?B\u65b9\u5411\u8fd0\u52a8\uff081\uff09\u51fa\u53d12\u79d2\u540e\uff0cAP=2\uff0cBQ=4\uff0c\u2234BP=8-2=6\uff0cPQ=BQ2+BP2=213\uff1b\uff083\u5206\uff09\uff082\uff09\u8bbe\u65f6\u95f4\u4e3at\uff0c\u5217\u65b9\u7a0b\u5f972t=8-1\u00d7t\uff0c\u89e3\u5f97t=83\uff1b\uff086\u5206\uff09\uff083\uff09\u5047\u8bbe\u76f4\u7ebfPQ\u80fd\u628a\u539f\u4e09\u89d2\u5f62\u5468\u957f\u5206\u6210\u76f8\u7b49\u7684\u4e24\u90e8\u5206\uff0c\u7531AB=8cm\uff0cBC=6cm\uff0c\u6839\u636e\u52fe\u80a1\u5b9a\u7406\u53ef\u77e5AC=10cm\uff0c\u5373\u4e09\u89d2\u5f62\u7684\u5468\u957f\u4e3a8+6+10=24cm\uff0c\u5219\u6709BP+BQ=12\u00d724=12\uff0c\u8bbe\u65f6\u95f4\u4e3at\uff0c\u5217\u65b9\u7a0b\u5f97\uff1a2t+\uff088-1\u00d7t\uff09=12\uff0c\u89e3\u5f97t=4\uff0c\u5f53t=4\u65f6\uff0c\u70b9Q\u8fd0\u52a8\u7684\u8def\u7a0b\u662f4\u00d72=8\uff1e6\uff0c\u6240\u4ee5\u76f4\u7ebfPQ\u4e0d\u80fd\u591f\u628a\u539f\u4e09\u89d2\u5f62\u5468\u957f\u5206\u6210\u76f8\u7b49\u7684\u4e24\u90e8\u5206\uff0e\uff0810\u5206\uff09
\u5f53\u70b9D\u4e0eB\u91cd\u5408\uff0c\u70b9E\u4e0eC\u91cd\u5408\u65f6\uff0cBE+ED\u6700\u5c0f=BC=5
①二秒后：BP=8-2=6
BQ=2*2=4
PQ=√6²+4²=2√13
②当t≤3时
BP=8-t，BQ=2t
8-t=2t，解得t=8/3
当t＞3时
AP=t，BP=8-t，CQ=2t-6，AQ=16-2t
PQ²=(16-2t)²+t²-2t(16-2t)*4/5
=8.2t²-64t+230.4
BQ²=36+(2t-6)²-2*6*(2t-6)*3/5
=4t²-38.4t+115.2
当PQ=BQ时，8.2t²-64t+230.4=4t²-38.4t+115.2，方程无解
当BP=PQ时，8.2t²-64t+230.4=（8-t)²，方程无解
当BP=BQ时，4t²-38.4t+115.2=（8-t)²，方程无解
综上可得当t=8/3时，△PQB为等腰三角形
③AP=t，BP=8-t，CQ=2t-6，AQ=16-2t
BQ²=36+(2t-6)²-2*6*(2t-6)*3/5
=4t²-38.4t+115.2
当BC=CQ，即2t-6=6，得t=6
当BQ=BC，即4t²-38.4t+115.2=6时，方程无解
当BQ=CQ，即4t²-38.4t+115.2=（2t-6)²，得t=5.5
综上可得，当t=5.5或t=6时，△BCQ为等腰三角形
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