题霸题霸学习平台
← 返回公开题库
九年级数学解答题一般
题目
如图,矩形ABCDABCD中,对角线ACAC,BDBD交于点OO,过点BBACAC的平行线,过点CCBDBD的平行线,这两条平行线交于点EE.
(1)(1)求证:四边形OBECOBEC是菱形;
(2)(2)AB=2AB=2,AD=23AD=2\sqrt{3},求菱形OBECOBEC的面积.
知识点:菱形的性质、矩形的判定与性质章节:第1章 特殊的平行四边形 / 1.2 矩形的性质与判定

答案与解析

答案

(1)(1)证明:BE\because BEAC,CEAC,CEBDBD
\therefore四边形OBECOBEC是平行四边形,
\because四边形ABCDABCD是矩形,
BD=AC\therefore BD=ACOB=12BDOB=\frac{1}{2}BDOC=12ACOC=\frac{1}{2}AC
OB=OC\therefore OB=OC
\therefore四边形OBECOBEC是菱形;
(2)(2)四边形ABCDABCD是矩形,
ABC=90\therefore \angle ABC=90^{\circ}BC=AD=23BC=AD=2\sqrt{3}OA=OCOA=OC
SABC=2SOBC\therefore S_{\triangle ABC}=2S_{\triangle OBC}
S菱形OBEC=2SOBC\because S_{菱形OBEC}=2S_{\triangle OBC}
S菱形OBEC=SABC=12ABBC=12×2×23=23\therefore S_{菱形OBEC}=S_{\triangle ABC}=\frac{1}{2}AB\cdot BC=\frac{1}{2}\times 2\times 2\sqrt{3}=2\sqrt{3}.

解析

(1)(1)证明:BE\because BEAC,CEAC,CEBDBD
\therefore四边形OBECOBEC是平行四边形,
\because四边形ABCDABCD是矩形,
BD=AC\therefore BD=ACOB=12BDOB=\frac{1}{2}BDOC=12ACOC=\frac{1}{2}AC
OB=OC\therefore OB=OC
\therefore四边形OBECOBEC是菱形;
(2)(2)四边形ABCDABCD是矩形,
ABC=90\therefore \angle ABC=90^{\circ}BC=AD=23BC=AD=2\sqrt{3}OA=OCOA=OC
SABC=2SOBC\therefore S_{\triangle ABC}=2S_{\triangle OBC}
S菱形OBEC=2SOBC\because S_{菱形OBEC}=2S_{\triangle OBC}
S菱形OBEC=SABC=12ABBC=12×2×23=23\therefore S_{菱形OBEC}=S_{\triangle ABC}=\frac{1}{2}AB\cdot BC=\frac{1}{2}\times 2\times 2\sqrt{3}=2\sqrt{3}.

自由组卷

围绕这道题再组一份练习 →

完整试卷

浏览同年级试卷结构 →