优优班--学霸训练营 > 题目详情
  • 如图\(①\),在\(\triangle ABC\)中,\(∠ACB=90^{\circ}\),\(∠B=30^{\circ}\),\(AC=1\),\(D\)为\(AB\)的中点,\(EF\)为\(\triangle ACD\)的中位线,四边形\(EFGH\)为\(\triangle ACD\)的内接矩形\((\)矩形的四个顶点均在\(\triangle ACD\)的边上\()\).
    \((1)\)计算矩形\(EFGH\)的面积;
    \((2)\)将矩形\(EFGH\)沿\(AB\)向右平移,\(F\)落在\(BC\)上时停止移动\(.\)在平移过程中,当矩形与\(\triangle CBD\)重叠部分的面积为\( \dfrac { \sqrt {3}}{16}\)时,求矩形平移的距离;
    \((3)\)如图\(③\),将\((2)\)中矩形平移停止时所得的矩形记为矩形\(E_{1}F_{1}G_{1}H_{1}\),将矩形\(E_{1}F_{1}G_{1}H_{1}\)绕\(G_{1}\)点按顺时针方向旋转,当\(H_{1}\)落在\(CD\)上时停止转动,旋转后的矩形记为矩形\(E_{2}F_{2}G_{1}H_{2}\),设旋转角为\(α\),求\(\cos α\)的值.
    【考点】矩形的性质,解直角三角形
    【分析】请登陆后查看
    【解答】请登陆后查看
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