已知A(0,2)为Y轴上一点,P为X轴上一动点,分别以AO,AP为边作等边三角形AOQ和APQ,问AOQB为梯形时P点的坐标.
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![已知A(0,2)为Y轴上一点,P为X轴上一动点,分别以AO,AP为边作等边三角形AOQ和APQ,问AOQB为梯形时P点的坐标.](/uploads/image/z/6335789-5-9.jpg?t=%E5%B7%B2%E7%9F%A5A%280%2C2%29%E4%B8%BAY%E8%BD%B4%E4%B8%8A%E4%B8%80%E7%82%B9%2CP%E4%B8%BAX%E8%BD%B4%E4%B8%8A%E4%B8%80%E5%8A%A8%E7%82%B9%2C%E5%88%86%E5%88%AB%E4%BB%A5AO%2CAP%E4%B8%BA%E8%BE%B9%E4%BD%9C%E7%AD%89%E8%BE%B9%E4%B8%89%E8%A7%92%E5%BD%A2AOQ%E5%92%8CAPQ%2C%E9%97%AEAOQB%E4%B8%BA%E6%A2%AF%E5%BD%A2%E6%97%B6P%E7%82%B9%E7%9A%84%E5%9D%90%E6%A0%87.)
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已知A(0,2)为Y轴上一点,P为X轴上一动点,分别以AO,AP为边作等边三角形AOQ和APQ,问AOQB为梯形时P点的坐标.
已知A(0,2)为Y轴上一点,P为X轴上一动点,分别以AO,AP为边作等边三角形AOQ和APQ,问AOQB为梯形时P点的
坐标.
已知A(0,2)为Y轴上一点,P为X轴上一动点,分别以AO,AP为边作等边三角形AOQ和APQ,问AOQB为梯形时P点的坐标.
①因为AOQB为梯形所以AQ平行与OP,∠QOP=90°-60°=30°,做QE垂直与X轴,所以2QE=OQ=OQ=2,QE=1,QE=根号内OQ²-QE²=根号3,设PE=X,则OP=根号3-X,AP=OA²+OP²=根号内4+(根号3-X)²,因为三角形AQP为正三角形,所以AP=QP,OQ=2QE=2,所以4+(根号3-X)²=4,(根号3-x)²=0,X1=X2=根号3.点P为(0,0)