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题目 #181110

UUID
fd87cdad-2303-4ad2-ac88-7cb52fdadb21
来源
沪教版/八下/周琼玟Q_18978751881_860.pdf 第 2 页,bbox [31.3, 142.25, 222.59, 223.91]
题型 / 学科
fill_blank / math / JUNIOR
知识树
KT_JUNIOR_MATH_STANDARD
难度
难度 2 / 5 简单 · 单一知识点
QA 状态
存疑 / 待修复 缺少答案 重试 0 次,得分 -
最终题 ID
-
错误
处理异常: LLM call failed after retries: HTTPConnectionPool(host='127.0.0.1', port=8092): Read timed out. (read timeout=120) [endpoints: http://127.0.0.1:8091/v1 -> http://127.0.0.1:8092/v1]

最终题干渲染(f_body 预览)

如图在, 中,ACB = 90, AC = 8,\(\frac{ABC}{以斜边} = 7\)为边向外作正方形\(\frac{BC}{BDE} \frac{AB}{CE}\)A ,连接CE,则 的长为() A .15 .16 .17 .18 B C D

答案 / 解析

答案(f_answer)缺少答案

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解析(f_ways)

如图所示过点作,EF ⊥ CA交\(E \frac{CA}{四边形}\frac{延长线}{ABDE}\)于, = ACB = 90, F F , =,BAE = 90,\(\frac{是正方形}{CAB} \frac{AE}{CBA} \frac{AB}{CAB}\) +  =  + FAE = 90, CBA = FAE,  (),\(\frac{CBA}{AC} \frac{FAEAAS}{CF}\) = BC = 7, = = 8, =\(\frac{AF}{AC} \frac{EF}{RtCEF} \frac{AF}{CE} + =15\)在  中由勾股定理得, =\(\sqrt{CF^{2}+EF^{2}} = \sqrt{15^{2}+8^{2}} =17\)故选 : . C

原图与资产(1)

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embedded_image2 [30.0, 236.37, 151.76, 358.77] 8908041b83ae0887…
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审计记录

阶段审计器结果得分依据(简短)
STRUCTURE structure-rule PASS 0.900
FORMULA formula-auditor PASS 1.000
KNOWLEDGE knowledge-auditor WARN -
VISION vision-auditor WARN 0.200

知识点(0)

无

事件

时间阶段级别消息
2026-11-10 04:43:46ERRORERROR处理异常: LLM call failed after retries: HTTPConnectionPool(host='127.0.0.1', port=8092): Read timed out. (read timeout=120) [endpoints: http://127.0.0.1:8091/v1 -> http://127.0.0.1:8092/v1]
2026-11-10 04:39:46FORMULAWARN公式转换未确认,保留原文本: solution
2026-11-10 04:35:46FORMULAWARN公式转换未确认,保留原文本: stem