金华十校2023-2024学年高一年级第二学期期末调研考试试题(数学)

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金华十校2023-2024学年高一年级第二学期期末调研考试试题(数学)试卷答案

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分析根据A,B,D三点共线,得出t+(2+t)=1,求出t的值,化简$\overrightarrow{CD}$=t$\frac{1}{2}$$\overrightarrow{CA}$+(2+t)$\overrightarrow{CB}$,得出$\overrightarrow{BD}$=$\frac{1}{2}$$\overrightarrow{BA}$,D是AB的中点,即可求出面积比是多少.

解答解:∵A,B,D三点共线,且$\overrightarrow{CD}$=t$\overrightarrow{CA}$+(2+t)$\overrightarrow{CB}$,
∴t+(2+t)=1,
解得t=-$\frac{1}{2}$;
∴$\overrightarrow{CD}$=-$\frac{1}{2}$$\overrightarrow{CA}$+$\frac{3}{2}$$\overrightarrow{CB}$,
∴$\overrightarrow{CD}$-$\overrightarrow{CB}$=$\frac{1}{2}$($\overrightarrow{CB}$-$\overrightarrow{CA}$),
即$\overrightarrow{BD}$=$\frac{1}{2}$$\overrightarrow{BA}$;如图所示,
∴BD=$\frac{1}{2}$AB,即BD=AD;
∴△CDB的面积和△CDA的面积之比为1:1.
故答案为:1:1.

点评本题考查了平面向量的应用问题,解题的关键是利用三点共线求出t的值,化简$\overrightarrow{CD}$=t$\frac{1}{2}$$\overrightarrow{CA}$+(2+t)$\overrightarrow{CB}$,得出D是AB的中点,是综合性题目.