如图,已知:△ABC中,AB=5,BC=3,AC=4,PQ//AB,P点在AC上(与点A、C不重合),Q点在BC上.(l)当△PQC的面积与四边形PABQ的面积相等时,求CP的长;(2)当△PQC的周长与四边形PABQ的周长相等时,求CP的
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![如图,已知:△ABC中,AB=5,BC=3,AC=4,PQ//AB,P点在AC上(与点A、C不重合),Q点在BC上.(l)当△PQC的面积与四边形PABQ的面积相等时,求CP的长;(2)当△PQC的周长与四边形PABQ的周长相等时,求CP的](/uploads/image/z/5265498-66-8.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%B7%B2%E7%9F%A5%EF%BC%9A%E2%96%B3ABC%E4%B8%AD%2CAB%EF%BC%9D5%2CBC%EF%BC%9D3%2CAC%EF%BC%9D4%2CPQ%2F%2FAB%2CP%E7%82%B9%E5%9C%A8AC%E4%B8%8A%EF%BC%88%E4%B8%8E%E7%82%B9A%E3%80%81C%E4%B8%8D%E9%87%8D%E5%90%88%EF%BC%89%2CQ%E7%82%B9%E5%9C%A8BC%E4%B8%8A%EF%BC%8E%EF%BC%88l%EF%BC%89%E5%BD%93%E2%96%B3PQC%E7%9A%84%E9%9D%A2%E7%A7%AF%E4%B8%8E%E5%9B%9B%E8%BE%B9%E5%BD%A2PABQ%E7%9A%84%E9%9D%A2%E7%A7%AF%E7%9B%B8%E7%AD%89%E6%97%B6%2C%E6%B1%82CP%E7%9A%84%E9%95%BF%EF%BC%9B%EF%BC%882%EF%BC%89%E5%BD%93%E2%96%B3PQC%E7%9A%84%E5%91%A8%E9%95%BF%E4%B8%8E%E5%9B%9B%E8%BE%B9%E5%BD%A2PABQ%E7%9A%84%E5%91%A8%E9%95%BF%E7%9B%B8%E7%AD%89%E6%97%B6%2C%E6%B1%82CP%E7%9A%84)
如图,已知:△ABC中,AB=5,BC=3,AC=4,PQ//AB,P点在AC上(与点A、C不重合),Q点在BC上.(l)当△PQC的面积与四边形PABQ的面积相等时,求CP的长;(2)当△PQC的周长与四边形PABQ的周长相等时,求CP的
如图,已知:△ABC中,AB=5,BC=3,AC=4,PQ//AB,P点在AC上(与点A、C不重合),Q点在BC上.(l)当△PQC的面积与四边形PABQ
的面积相等时,求CP的长;(2)当△PQC的周长与四边形PABQ的周长相等时,求CP的长;(3)试问:在AB上是否存在点M,使得△PQM为等腰直角三角形?若不存在,请简要说明理由;若存在,请求出PQ的长.
如图,已知:△ABC中,AB=5,BC=3,AC=4,PQ//AB,P点在AC上(与点A、C不重合),Q点在BC上.(l)当△PQC的面积与四边形PABQ的面积相等时,求CP的长;(2)当△PQC的周长与四边形PABQ的周长相等时,求CP的
(1)由S△ABC=1/2×3×4=6,
∴S△PQC=6÷2=3,
由PQ‖AB,AC=4,BC=3,
设PC=4x,QC=3x,
得:1/2×4x×3x=3,
x=±√2/2,(x=-√2/2舍去)
∴x=√2/2,即PC=2√2,QC=3√2/2.
(2)设PC=4t,QC=3t,PQ=5t,
PA=4-4t,BQ=3-3t,
∴3t+4t+5t=(3-3t)+(4-4t)+5t+5
t=6/7
∴PC=4t=24/7.
(3)符合条件的M点是存在的.
①设QM=PM=x,∠OMQ=90°,
BQ=3-x,PA=4-x,
由QM‖AC,
∴(3-x)/x=x/(4-x),
12-7x+x²=x²,
∴x=12/7.
∴PQ²=(12/7)²+(12/7)²
PQ²=288/49,
∴PQ=12√2/7.
②设PQ=QM=5x,∠MQP=90°,
QC=3x,PC=4x,
由△BMQ∽△QCP,
∴(3-3x)/5x=5x/4x,
x=12/37.
∴PQ=5x=60/37.
③设PQ=PM=5x,∠MPQ=90°,
QC=3x,PC=4x,PA=4-4x,
由△PQC∽△APM,
∴3x/5x=5x/(4-4x)
x=12/37,
∴PQ=5x=60/37.