想必現(xiàn)在有很多小伙伴對于如圖,直線$y=- \dfrac{1}{2}x+2$與$x$軸交于$C$,與$y$軸交于$D$,以$CD$為邊作矩形$CDAB$,點$A$在$x$軸上,雙曲線$y= \dfrac{k}{x}\left(k \lt 0\right)$經(jīng)過點$B$與直線$CD$交于$E$,$EM\bot x$軸于$M$,則$S_{四邊形BEMC}=$___.","title_text":"如圖,直線$y=- \dfrac{1}{2}x+2$與$x$軸交于$C$,與$y$軸交于$D$,以$CD$為邊作矩形$CDAB$,點$A$在$x$軸上,雙曲線$y= \dfrac{k}{x}\left(k \lt 0\right)$經(jīng)過點$B$與直線$CD$交于$E$,$EM\bot x$軸于$M$,則$S_{四邊形BEMC}=$___.方面的知識都比較想要了解,那么今天小好小編就為大家收集了一些關(guān)于如圖,直線$y=- \dfrac{1}{2}x+2$與$x$軸交于$C$,與$y$軸交于$D$,以$CD$為邊作矩形$CDAB$,點$A$在$x$軸上,雙曲線$y= \dfrac{k}{x}\left(k \lt 0\right)$經(jīng)過點$B$與直線$CD$交于$E$,$EM\bot x$軸于$M$,則$S_{四邊形BEMC}=$___.","title_text":"如圖,直線$y=- \dfrac{1}{2}x+2$與$x$軸交于$C$,與$y$軸交于$D$,以$CD$為邊作矩形$CDAB$,點$A$在$x$軸上,雙曲線$y= \dfrac{k}{x}\left(k \lt 0\right)$經(jīng)過點$B$與直線$CD$交于$E$,$EM\bot x$軸于$M$,則$S_{四邊形BEMC}=$___.方面的知識分享給大家,希望大家會喜歡哦。
根據(jù)題意,直線$y=-dfrac{1}{2}x+2$與$x$軸交于$C$,與$y$軸交于$D$,
分別令$x=0$,$y=0$,
得$y=2$,$x=4$,
即$Dleft(0,2right)$,$Cleft(4,0right)$,
即$DC=2sqrt {5}$,
又$ADbot DC$且過點$D$,
所以直線$AD$所在函數(shù)解析式為:$y=2x+2$,
令$y=0$,得$x=-1$,
即$Aleft(-1,0right)$,
同理可得$B$點的坐標為$Bleft(3,-2right)$
又$B$為雙曲線$y=dfrac{k}{x}left(k lt 0right)$上,
代入得$k=-6$.
即雙曲線的解析式為$y=dfrac{-6}{x}$
與直線$DC$聯(lián)立$,$
$left{begin{array}{}y=dfrac{-6}{x} y=-dfrac{1}{2}x+2end{array}right.$,
得$left{begin{array}{}x=6 y=-1end{array}right.$和$left{begin{array}{}x=-2 y=3end{array}right.$
根據(jù)題意,$left{begin{array}{}x=-2 y=3end{array}right.$不合題意,
故點$E$的坐標為$left(6,-1right)$.
所以$BC=sqrt {5}$,$CE=sqrt {5}$,
$CM=2$,$EM=1$,
所以$S_{triangle BEC}=dfrac{1}{2}times BCtimes EC=dfrac{5}{2}$,
$S_{triangle EMC}=dfrac{1}{2}times EMtimes CM=1$,
故$S_{四BEMC}=S_{triangle BEC}+S_{triangle EMC}=dfrac{7}{2}$.
故答案為:$dfrac{7}{2}$.
本文到此結(jié)束,希望對大家有所幫助。