勾股定理是人类早期发现并证明的重要数学定理之一,是数形结合的重要纽带.数学家欧几里得利用下图验证了勾股定理.以直角三角形
的三条边为边长向外作正方形
,正方形
,正方形
,连接
,
,过点C作
于点J,交
于点K.设正方形
的面积为
,正方形
的面积为
,矩形
的面积为
,矩形
的面积为
,下列结论中:①
;②
;③
;④
,正确的结论有( )
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/7bef5239ddbb0972700ce01daf9ee7cf.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/0c35e7a990d8ff3e55945fd4718a9740.png)
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A.1个 | B.2个 | C.3个 | D.4个 |
21-22八年级下·福建福州·期中 查看更多[8]
福建省福州市晋安区2021-2022学年八年级下学期期中数学试题(已下线)3.1 勾股定理(培优分阶练)-2022-2023学年八年级数学上册课后培优分级练(苏科版)(已下线)期中押题预测卷(考试范围:第一~四章)-【帮课堂】2022-2023学年八年级数学上册同步精品讲义(北师大版)(已下线)专题1.5 勾股定理章末拔尖卷-2023-2024学年八年级数学上册举一反三系列(北师大版)(已下线)专题3.5 勾股定理章末拔尖卷-2023-2024学年八年级数学上册举一反三系列(苏科版)(已下线)专题14.5 勾股定理章末拔尖卷-2023-2024学年八年级数学上册举一反三系列(华东师大版)福建省莆田市城厢区莆田哲理中学2023-2024学年八年级下学期月考数学试题福建省福州第三十二中学2023-2024学年八年级下学期月考数学试题
更新时间:2024-04-01 20:15:20
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【推荐1】如图,大正方形中有2个小正方形,这两个小正方形的面积分别是
和
,则
的值是( )
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/e097c8d4c948de063796bd19f85b3a9a.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/1e0bd63f55069a3bc870915010b39225.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/235f0a6fb218d28383e6f27f2df1f50f.png)
A.![]() | B.![]() | C.1 | D.![]() |
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【推荐2】如图,在
中,以AC为直角边向外作
,分别以AB,BC,CD,DA为直径向外作半圆,面积分别记为S1,S2,S3,S4,已知
,
,
,则S4为( )
![](https://img.xkw.com/dksih/QBM/editorImg/2022/10/4/18a96bbf-fa81-4ef5-8cfa-7ef17884933e.png?resizew=138)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/dd967903ed5a6f640a5b801ec8be0070.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/d41225b0c6ff901aef01d5094310b82b.png)
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![](https://img.xkw.com/dksih/QBM/editorImg/2022/10/4/18a96bbf-fa81-4ef5-8cfa-7ef17884933e.png?resizew=138)
A.2 | B.3 | C.![]() | D.![]() |
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名校
【推荐1】如图,在矩形
中,
,
,
于
,
于
,
平分
交
于点
,交
延长线与点
,则下列说法中正确的有( )个
①
;②
;③
;④
;⑤![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b7ff55322db571cf7d02569a738b5798.png)
![](https://img.xkw.com/dksih/QBM/2020/12/9/2615411801415680/2621737074024448/STEM/07487de6-9fe4-49da-91a9-60046046f386.png?resizew=241)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/411b38a18046fea8e9fab1f9f9b80a5f.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/efc6e4b936d7a800e839a30c3839574d.png)
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![](https://staticzujuan.xkw.com/quesimg/Upload/formula/1fc56c77464a17a1e97b568762a3e2c6.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/a0ed1ec316bc54c37c4286c208f55667.png)
①
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/1596f7c0d0bab4e7e586d705d323e0d8.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/7a5a7456be827d9218aa7a2087ed1fa5.png)
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A.2 | B.3 | C.4 | D.5 |
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【推荐2】如图,在正方形ABCD中,E,F分别为BC、CD的中点,连接AE,BF交于点G,将△BCF沿BF对折,得到△BPF,延长FP交BA延长线于点Q,下列结论:①
;②
;③
;④
;⑤
.其中正确的结论有( )
![](https://img.xkw.com/dksih/QBM/editorImg/2022/9/14/b281c839-1e1b-4b7f-9d0c-db37dea9d5a5.png?resizew=195)
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