如图20,在平面直角坐标系中,四边形OABC是矩形,点B的坐标为(4,3).平行于对角线AC的直线m从原点O出发
如图20,在平面直角坐标系中,四边形OABC是矩形,点B的坐标为(4,3).平行于对角线AC的直线m从原点O出发,沿x轴正方向以每秒1个单位长度的速度运动,设直线m与矩形...
如图20,在平面直角坐标系中,四边形OABC是矩形,点B的坐标为(4,3).平行于对角线AC的直线m从原点O出发,沿x轴正方向以每秒1个单位长度的速度运动,设直线m与矩形OABC的两边分别交于点M、N,直线m运动的时间为t(秒). 图20 (1) 点A的坐标是__________,点C的坐标是__________; (2) 当t= 秒或 秒时,MN= 0.5AC; (3) 设△OMN的面积为S,求S与t的函数关系式; (4) 探求(3)中得到的函数S有没有最大值?若有,求出最大值; 若没有,请说明理由。
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解:(1)(4,0),(0,3); ·················································································· 2分
(2) 2,6; ········································································································· 4分
(3) 当0<t≤4时,OM=t.
由△OMN∽△OAC,得,
∴ ON=,S=. ···································· 6分
当4<t<8时,
如图,∵ OD=t,∴ AD= t-4.
方法一:
由△DAM∽△AOC,可得AM=,∴ BM=6-. ··························· 7分
由△BMN∽△BAC,可得BN==8-t,∴ CN=t-4. ·································· 8分
S=矩形OABC的面积-Rt△OAM的面积- Rt△MBN的面积- Rt△NCO的面积
=12--(8-t)(6-)-
=. ··························································································· 10分
方法二:
易知四边形ADNC是平行四边形,∴ CN=AD=t-4,BN=8-t.·································· 7分
由△BMN∽△BAC,可得BM==6-,∴ AM=.······ 8分
以下同方法一.
(4) 有最大值.
方法一:
当0<t≤4时,
∵ 抛物线S=的开口向上,在对称轴t=0的右边, S随t的增大而增大,
∴ 当t=4时,S可取到最大值=6; ················ 11分
当4<t<8时,
∵ 抛物线S=的开口向下,它的顶点是(4,6),∴ S<6.
综上,当t=4时,S有最大值6. ······································································· 12分
方法二:
∵ S=
∴ 当0<t<8时,画出S与t的函数关系图像,如图所示. ······························ 11分
显然,当t=4时,S有最大值6
(2) 2,6; ········································································································· 4分
(3) 当0<t≤4时,OM=t.
由△OMN∽△OAC,得,
∴ ON=,S=. ···································· 6分
当4<t<8时,
如图,∵ OD=t,∴ AD= t-4.
方法一:
由△DAM∽△AOC,可得AM=,∴ BM=6-. ··························· 7分
由△BMN∽△BAC,可得BN==8-t,∴ CN=t-4. ·································· 8分
S=矩形OABC的面积-Rt△OAM的面积- Rt△MBN的面积- Rt△NCO的面积
=12--(8-t)(6-)-
=. ··························································································· 10分
方法二:
易知四边形ADNC是平行四边形,∴ CN=AD=t-4,BN=8-t.·································· 7分
由△BMN∽△BAC,可得BM==6-,∴ AM=.······ 8分
以下同方法一.
(4) 有最大值.
方法一:
当0<t≤4时,
∵ 抛物线S=的开口向上,在对称轴t=0的右边, S随t的增大而增大,
∴ 当t=4时,S可取到最大值=6; ················ 11分
当4<t<8时,
∵ 抛物线S=的开口向下,它的顶点是(4,6),∴ S<6.
综上,当t=4时,S有最大值6. ······································································· 12分
方法二:
∵ S=
∴ 当0<t<8时,画出S与t的函数关系图像,如图所示. ······························ 11分
显然,当t=4时,S有最大值6
2012-04-28
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解
②当0<t≤4时,OM=t,由△OMN∽△OAC,得 ,∴ON= t,S= =
∴t=2;当4<t<8时,如图,∵OD=t,∴AD=t-4.
由△DAM∽△AOC,可得AM= (t-4)∴BM=6- t.
由△BMN∽△BAC,可得BN= BM=8-t∴CN=t-4
S=矩形OABC的面积-Rt△OAM的面积-Rt△MBN的面积-Rt△NCO的面积
=12- - (8-t)(6- t)- =- +3t,∴- +3t=
解得 取t=4+2 故当t=2或4+2 时,△OMN的面积S= .
②当0<t≤4时,OM=t,由△OMN∽△OAC,得 ,∴ON= t,S= =
∴t=2;当4<t<8时,如图,∵OD=t,∴AD=t-4.
由△DAM∽△AOC,可得AM= (t-4)∴BM=6- t.
由△BMN∽△BAC,可得BN= BM=8-t∴CN=t-4
S=矩形OABC的面积-Rt△OAM的面积-Rt△MBN的面积-Rt△NCO的面积
=12- - (8-t)(6- t)- =- +3t,∴- +3t=
解得 取t=4+2 故当t=2或4+2 时,△OMN的面积S= .
参考资料: 网上
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解:(1)(4,0),(0,3); ·················································································· 2分
(2) 2,6; ········································································································· 4分
(3) 当0<t≤4时,OM=t.
由△OMN∽△OAC,得,
∴ ON=,S=. ···································· 6分
当4<t<8时,
如图,∵ OD=t,∴ AD= t-4.
方法一:
由△DAM∽△AOC,可得AM=,∴ BM=6-. ··························· 7分
由△BMN∽△BAC,可得BN==8-t,∴ CN=t-4. ·································· 8分
S=矩形OABC的面积-Rt△OAM的面积- Rt△MBN的面积- Rt△NCO的面积
=12--(8-t)(6-)-
=. ··························································································· 10分
方法二:
易知四边形ADNC是平行四边形,∴ CN=AD=t-4,BN=8-t.·································· 7分
由△BMN∽△BAC,可得BM==6-,∴ AM=.······ 8分
以下同方法一.
(4) 有最大值.
方法一:
当0<t≤4时,
∵ 抛物线S=的开口向上,在对称轴t=0的右边, S随t的增大而增大,
∴ 当t=4时,S可取到最大值=6; ················ 11分
当4<t<8时,
∵ 抛物线S=的开口向下,它的顶点是(4,6),∴ S<6.
综上,当t=4时,S有最大值6. ······································································· 12分
方法二:
∵ S=
∴ 当0<t<8时,画出S与t的函数关系图像,如图所示. ······························ 11分
显然,当t=4时,S有最大值6. ··································································· 12分
(2) 2,6; ········································································································· 4分
(3) 当0<t≤4时,OM=t.
由△OMN∽△OAC,得,
∴ ON=,S=. ···································· 6分
当4<t<8时,
如图,∵ OD=t,∴ AD= t-4.
方法一:
由△DAM∽△AOC,可得AM=,∴ BM=6-. ··························· 7分
由△BMN∽△BAC,可得BN==8-t,∴ CN=t-4. ·································· 8分
S=矩形OABC的面积-Rt△OAM的面积- Rt△MBN的面积- Rt△NCO的面积
=12--(8-t)(6-)-
=. ··························································································· 10分
方法二:
易知四边形ADNC是平行四边形,∴ CN=AD=t-4,BN=8-t.·································· 7分
由△BMN∽△BAC,可得BM==6-,∴ AM=.······ 8分
以下同方法一.
(4) 有最大值.
方法一:
当0<t≤4时,
∵ 抛物线S=的开口向上,在对称轴t=0的右边, S随t的增大而增大,
∴ 当t=4时,S可取到最大值=6; ················ 11分
当4<t<8时,
∵ 抛物线S=的开口向下,它的顶点是(4,6),∴ S<6.
综上,当t=4时,S有最大值6. ······································································· 12分
方法二:
∵ S=
∴ 当0<t<8时,画出S与t的函数关系图像,如图所示. ······························ 11分
显然,当t=4时,S有最大值6. ··································································· 12分
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两个函数解析式要分类说明第三问
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2011-04-10
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如图20,在平面直角坐标系中,四边形OABC是矩形,点B的坐标为(4,3).平行于对角线AC的直线m从原点O出发 悬赏分:0 | 离问题结束还有 17 小时 | 提问者:10602787566
如图20,在平面直角坐标系中,四边形OABC是矩形,点B的坐标为(4,3).平行于对角线AC的直线m从原点O出发,沿x轴正方向以每秒1个单位长度的速度运动,设直线m与矩形OABC的两边分别交于点M、N,直线m运动的时间为t(秒). 图20 (1) 点A的坐标是__________,点C的坐标是__________; (2) 当t= 秒或 秒时,MN= 0.5AC; (3) 设△OMN的面积为S,求S与t的函数关系式; (4) 探求(3)中得到的函数S有没有最大值?若有,求出最大值; 若没有,请说明理由。
如图20,在平面直角坐标系中,四边形OABC是矩形,点B的坐标为(4,3).平行于对角线AC的直线m从原点O出发,沿x轴正方向以每秒1个单位长度的速度运动,设直线m与矩形OABC的两边分别交于点M、N,直线m运动的时间为t(秒). 图20 (1) 点A的坐标是__________,点C的坐标是__________; (2) 当t= 秒或 秒时,MN= 0.5AC; (3) 设△OMN的面积为S,求S与t的函数关系式; (4) 探求(3)中得到的函数S有没有最大值?若有,求出最大值; 若没有,请说明理由。
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