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九年级数学选择题一般
题目
20232023个边长为11的正方形按如图所示的方式排列,点AA,A1A_{1},A2A_{2},A3A_{3},A2022\ldots A_{2022}和点MM,M1M_{1},M2M_{2},\ldots,M2021M_{2021}是正方形的顶点,连接AM1AM_{1},AM2AM_{2},AM3AM_{3},AM2021\ldots AM_{2021}分别交正方形的边A1MA_{1}M,A2M1A_{2}M_{1},A3M2A_{3}M_{2},A2022M2021\ldots A_{2022}M_{2021}于点N1N_{1},N2N_{2},N3N2022N_{3}\ldots N_{2022},四边形M1N1A1A2M_{1}N_{1}A_{1}A_{2}的面积是S1S_{1},四边形M2N2A2A3M_{2}N_{2}A_{2}A_{3}的面积是S2S_{2},\ldots,则S2022S_{2022}为( )
A.
20212022\frac{2021}{2022}
B.
20222023\frac{2022}{2023}
C.
40434044\frac{4043}{4044}
D.
40454046\frac{4045}{4046}
知识点:正方形的性质、相似三角形的性质I、相似三角形的判定与性质章节:第24章 相似三角形 / 第3节 相似三角形 / 24.5 相似三角形的性质

答案与解析

答案

D

解析

如图,

根据题意得:MN1MN_{1}OAOA
M1MN1\therefore \triangle M_{1}MN_{1}M1OA\triangle M_{1}OA
MM1OM1=MN1AO=12\therefore \frac{M{M_1}}{O{M_1}}=\frac{M{N_1}}{AO}=\frac{1}{2}
\therefore四边形M1N1A1A2M_{1}N_{1}A_{1}A_{2}的面积为S1=112×1×12=114=34{S_1}=1-\frac{1}{2}×1×\frac{1}{2}=1-\frac{1}{4}=\frac{3}{4}
同理:M1M2OM2=M1N2OA=13\frac{{M_1}{M_2}}{O{M_2}}=\frac{{M_1}{N_2}}{OA}=\frac{1}{3}
\therefore四边形M2N2A2A3M_{2}N_{2}A_{2}A_{3}的面积为S2=112×1×13=116=56{S_2}=1-\frac{1}{2}×1×\frac{1}{3}=1-\frac{1}{6}=\frac{5}{6}
\ldots \ldots
\therefore四边形MnNnAnAn+1M_{n}N_{n}A_{n}A_{n+1}的面积为Sn=112(n+1)=2n+12n+2{S_n}=1-\frac{1}{2(n+1)}=\frac{2n+1}{2n+2}
S2022=112(2022+1)=40454046\therefore {S_{2022}}=1-\frac{1}{2(2022+1)}=\frac{4045}{4046}.
故选:DD.

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