Question

平面直角坐标系中，四边形OABC是矩形，点B的坐标（4，3）。平行于对角线AC的直线m从原点O出发，沿

平面直角坐标系中，四边形OABC是矩形，点B的坐标（4，3）。平行于对角线AC的直线m从原点O出发，沿
看图吧，双击打开，谢谢

Answer 1

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：州绝

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=．伍颂······ 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=的腔迹郑开口向上，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

Answer 2

(1)A(4,0) C(0,3) (2)当t=2秒或6秒时，MN=1/洞槐正2AC（理由是，此时MN是AC的中位线）
（3）明做S＝1/2×t×(3/4t)=3/纳悔8t平方（t<=4)或S=1/2×3×t-1/2(t-4)×3/4(t-4)=-3/8t平方＋9/2t-6(t>4)

Answer 3

我也正在找这题

Answer 4

就是这样

Answer 5

到网上查嘛

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