关于证明题,高手进来救救偶吖,如图,已知;△ABC中,AB=5,BC=3,AC=4,PQ//AB,P点在AC上,Q点在BC上.△PQC相似于△ABC(1)当△PQC的面积与四边形PABQ的面积相等时,求CP的上(2)当△PQC的周长与四边形PABQ的周长相
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![关于证明题,高手进来救救偶吖,如图,已知;△ABC中,AB=5,BC=3,AC=4,PQ//AB,P点在AC上,Q点在BC上.△PQC相似于△ABC(1)当△PQC的面积与四边形PABQ的面积相等时,求CP的上(2)当△PQC的周长与四边形PABQ的周长相](/uploads/image/z/5372241-33-1.jpg?t=%E5%85%B3%E4%BA%8E%E8%AF%81%E6%98%8E%E9%A2%98%2C%E9%AB%98%E6%89%8B%E8%BF%9B%E6%9D%A5%E6%95%91%E6%95%91%E5%81%B6%E5%90%96%2C%E5%A6%82%E5%9B%BE%2C%E5%B7%B2%E7%9F%A5%3B%E2%96%B3ABC%E4%B8%AD%2CAB%3D5%2CBC%3D3%2CAC%3D4%2CPQ%2F%2FAB%2CP%E7%82%B9%E5%9C%A8AC%E4%B8%8A%2CQ%E7%82%B9%E5%9C%A8BC%E4%B8%8A.%E2%96%B3PQC%E7%9B%B8%E4%BC%BC%E4%BA%8E%E2%96%B3ABC%281%29%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%E4%B8%8A%282%29%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)
关于证明题,高手进来救救偶吖,如图,已知;△ABC中,AB=5,BC=3,AC=4,PQ//AB,P点在AC上,Q点在BC上.△PQC相似于△ABC(1)当△PQC的面积与四边形PABQ的面积相等时,求CP的上(2)当△PQC的周长与四边形PABQ的周长相
关于证明题,高手进来救救偶吖,
如图,已知;△ABC中,AB=5,BC=3,AC=4,PQ//AB,P点在AC上,Q点在BC上.△PQC相似于△ABC
(1)当△PQC的面积与四边形PABQ的面积相等时,求CP的上
(2)当△PQC的周长与四边形PABQ的周长相等时,求CP的长.
图在这
关于证明题,高手进来救救偶吖,如图,已知;△ABC中,AB=5,BC=3,AC=4,PQ//AB,P点在AC上,Q点在BC上.△PQC相似于△ABC(1)当△PQC的面积与四边形PABQ的面积相等时,求CP的上(2)当△PQC的周长与四边形PABQ的周长相
(1)△PQC相似于△ABC
∴CP/CA=CQ/CB=PQ/AB
∴CQ=(CP/CA)*CB=(3/4)CP;PQ=(CP/CA)*AB=(5/4)CP
由AC^2+BC^2=AB^2
△ABC为直角三角形.∠C=90°
S△ABC=(1/2)*AC*BC=(1/2)*3*4=6
S△PQC=(1/2)*CP*CQ=(1/2)*CP*(3/4)CP=(3/8)CP^2
四边形PABQ面积=S△ABC-S△PQC=S△PQC
∴S△PQC=(1/2)S△ABC
(3/8)CP^2=3
∴CP=2√2
(2)△PQC的周长=CP+CQ+PQ=CP+(3/4)CP+(5/4)CP=3CP
四边形PABQ的周长=PQ+AP+AB+BQ=(5/4)CP+(AC-CP)+5+(BC-CQ)=(5/4)CP+(4-CP)+5+(3-(3/4)CP)=12-(1/2)CP
根据题意,3CP=12-(1/2)CP
∴CP=24/7
(1)角C=90度
△PQC的面积是△ABC的面积的1/2
即(CP×CQ)/(CA×CB)=1/2
而CP/CA=CQ/CB
所以(CP/AC)^2=1/2
(2)因为△PQC的周长与四边形PABQ的周长相等
所以CP+CQ=(AC+BC+AB)/2=6
又CP/AC=CQ/CB
即CQ=3/4CP
故CP=24/7