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九年级数学填空题一般
题目
如图,在梯形ABCDABCD,AB,ABCDCD,且AB:CD=3:2AB:CD=3:2,点EE是边CDCD的中点,联结BEBE交对角线ACAC于点FF,若AB=a\overrightarrow{AB}=\overrightarrow{a},AD=b\overrightarrow{AD}=\overrightarrow{b}.
(1)(1)直接用a\overrightarrow{a}b\overrightarrow{b}表示DC=\overrightarrow{DC}=______;AC=\overrightarrow{AC}=______;AF=\overrightarrow{AF}=______;
(2)(2)求作BF\overrightarrow{BF}BA\overrightarrow{BA}BC\overrightarrow{BC}方向上的分向量.
((不要求写作法,但要保留作图痕迹,并指出所作图中表示结论的分向量)
知识点:平面向量I章节:第22章 四边形 / 第4节 平面向量及其加减运算 / 22.7 平面向量

答案与解析

答案

(1)AB\left(1\right)\because ABCDCDAB:CD=3:2AB:CD=3:2
CD=23AB\therefore CD=\frac{2}{3}AB
AB=a^\because \overrightarrow{AB}=\hat{a}DC=23a\therefore \overrightarrow{DC}=\frac{2}{3}\overrightarrow{a}AC=AD+DC=b+23a\overrightarrow{AC}=\overrightarrow{AD}+\overrightarrow{DC}=\overrightarrow{b}+\frac{2}{3}\overrightarrow{a}
E\because ECDCD的中点,
EC:AB=1:3\therefore EC:AB=1:3
EC\because ECABAB
AF:FC\therefore AF:FCAB:CE=3:1AB:CE=3:1
AF=34AC\therefore AF=\frac{3}{4}AC
AF=34(b+23a)=34b+12a\therefore \overrightarrow{AF}=\frac{3}{4}(\overrightarrow{b}+\frac{2}{3}\overrightarrow{a})=\frac{3}{4}\overrightarrow{b}+\frac{1}{2}\overrightarrow{a}.
故答案为:23a\frac{2}{3}\overrightarrow{a}b+23a\overrightarrow{b}+\frac{2}{3}\overrightarrow{a}34b+12a\frac{3}{4}\overrightarrow{b}+\frac{1}{2}\overrightarrow{a}.
(2)(2)如图,BM\overrightarrow{BM}BN\overrightarrow{BN}即为所求.

解析

(1)AB\left(1\right)\because ABCDCDAB:CD=3:2AB:CD=3:2
CD=23AB\therefore CD=\frac{2}{3}AB
AB=a^\because \overrightarrow{AB}=\hat{a}DC=23a\therefore \overrightarrow{DC}=\frac{2}{3}\overrightarrow{a}AC=AD+DC=b+23a\overrightarrow{AC}=\overrightarrow{AD}+\overrightarrow{DC}=\overrightarrow{b}+\frac{2}{3}\overrightarrow{a}
E\because ECDCD的中点,
EC:AB=1:3\therefore EC:AB=1:3
EC\because ECABAB
AF:FC\therefore AF:FCAB:CE=3:1AB:CE=3:1
AF=34AC\therefore AF=\frac{3}{4}AC
AF=34(b+23a)=34b+12a\therefore \overrightarrow{AF}=\frac{3}{4}(\overrightarrow{b}+\frac{2}{3}\overrightarrow{a})=\frac{3}{4}\overrightarrow{b}+\frac{1}{2}\overrightarrow{a}.
故答案为:23a\frac{2}{3}\overrightarrow{a}b+23a\overrightarrow{b}+\frac{2}{3}\overrightarrow{a}34b+12a\frac{3}{4}\overrightarrow{b}+\frac{1}{2}\overrightarrow{a}.
(2)(2)如图,BM\overrightarrow{BM}BN\overrightarrow{BN}即为所求.

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