25.如图.将含角的直角三角板()绕其直角顶点逆时针旋转解().得到.与相交于点.过点作交于点.连结.设.的面积为.的面积为. (1)求证:是直角三角形, (2)试求用表示的函数关系式.并写出的取值范围, (3)以点为圆心.为半径作. ①当直线与相切时.试探求与之间的关系, ②当时.试判断直线与的位置关系.并说明理由. 25. (1). 又.································································· 1分 又.···························································· 2分 . .即是直角三角形,······························································ 3分 (2)在中.. . .. ,··············································································· 4分 ,··························· 5分 (3)①直线与相切时.则. . . .································································· 6分 又. 是等边三角形.. . 又,······································································ 7分 ②当时. 则有.解之得或,··················································· 8分 (i)当时.. 在中... 在中..········································· 9分 .即. 直线与相离,···························································································· 10分 (ii)当时. 同理可求出:.·············································· 11分 . 直线与相交.···························································································· 12分 查看更多

 

题目列表(包括答案和解析)

(1)已知:有两块完全相同的含45°角的三角板,如图,将Rt△DEF的直角的=顶点D放在Rt△ABC斜边AB的中点处,这时两块三角板重叠部分△DBC的面积是△ABC的面积的________;

(2)如图,点D不动,将Rt△DEF绕着顶点D旋转α(0°<∠α<90°),这时两块三角板重叠部分为任意四边形DNCM,这时四边形DNCM的面积是△ABC的面积的________;

(3)若Rt△DEF的顶点D在AB上移动(不与点A、B重合),且两条直角边与Rt△ABC的两条直角边相交,是否存在一点,使得两块三角板重叠部分的面积是Rt△ABC的面积的,如果存在,请在图中画出此时的图形,并说明点D在AB上的位置.如果不存在,说明理由.

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(1)已知:有两块完全相同的含45°角的三角板,如图,将Rt△DEF的直角的=顶点D放在Rt△ABC斜边AB的中点处,这时两块三角板重叠部分△DBC的面积是△ABC的面积的________

(2)如图,点D不动,将Rt△DEF绕着顶点D旋转α(0°<∠α<90°),这时两块三角板重叠部分为任意四边形DNCM,这时四边形DNCM的面积是△ABC的面积的________

(3)若Rt△DEF的顶点D在AB上移动(不与点A、B重合),且两条直角边与Rt△ABC的两条直角边相交,是否存在一点,使得两块三角板重叠部分的面积是Rt△ABC的面积的,如果存在,请在图中画出此时的图形,并说明点D在AB上的位置.如果不存在,说明理由.

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如图,30°角的直角三角板ABC,∠ACB=90°,短边BC=1cm,将Rt△ABC在直线上绕三角形右下角的顶点顺时针翻转1次,点A经过的路径长为    ;顺时针连续翻转3次,点A经过的路径长为    ;顺时针连续翻转3n次,点A经过的路径长为    ;顺时针连续翻转3n+1次,点A经过的路径长为   

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如图用一个有角的直角三角板估测学校旗杆AB的高度.他将角的直角边水平放在1.5米高的支架CD上,三角板的斜边与旗杆的顶点在同一直线上,他又量得D、B的距离为5米,则旗杆AB的高度约为________米(精确到1米,取1.73)

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15、如图,把含30°的直角三角板ABC绕直角顶点B顺时钟旋转到△DBE的位置,AC交BD于点G,BC交DE于点F,若∠BFE=85°,则∠AGD的度数是
65°

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