{"ok":true,"question":{"id":22440,"question_uuid":"0db60c21-9484-43c8-8268-e05d5d6549de","job_id":3,"source_file":"人教版/七下/2026江苏七年级期末数学试卷+答案695.pdf","source_page_start":3,"source_page_end":3,"source_bbox":"[30.22, 380.32, 418.58, 462.34]","subject_code":"math","grade_level":"SENIOR","tree_code":"KT_SENIOR_MATH_STANDARD","question_type":"comprehensive","canonical_json":"{\"year\": null, \"blocks\": [{\"text\": \"如图，在平面直角坐标系中，已知 △ ABC的三个顶点坐标分别是 A(3, −1), B(1, −4), C(3, −3)，将△ABC平移后得到△A′B′C′，其中点 A, B, C的对应点分别为A′, B′, C′．已知点A′的坐标是A′(2, 2)．\", \"type\": \"text\"}, {\"kind\": \"embedded_image\", \"type\": \"figure\", \"width\": 520, \"height\": 430, \"sha256\": \"828f6adb3fb393378faf94a6c29fa773245544610105cb0f4d49addaf67bc619\", \"caption\": \"\", \"asset_key\": \"math/2026/09/2b06312b-7e3f-494e-ae3e-21d10207fcf7/0db60c21-9484-43c8-8268-e05d5d6549de/828f6adb3fb3.png\", \"public_url\": \"http://tikupdfpng.mtwlkj.net:5129/qbank/math/2026/09/2b06312b-7e3f-494e-ae3e-21d10207fcf7/0db60c21-9484-43c8-8268-e05d5d6549de/828f6adb3fb3.png\", \"source_bbox\": [30.21787452697754, 462.8291015625, 126.6787567138672, 553.1654663085938], \"source_page\": 3}, {\"type\": \"subquestion\", \"index\": 1, \"blocks\": [{\"text\": \"①画出△A′B′C′； ②△ABC上任一点D(x, y)平移后对应的点为D′(x + a, y + b)，则ab =____ ；\", \"type\": \"text\"}]}, {\"type\": \"subquestion\", \"index\": 2, \"blocks\": [{\"text\": \"求△B′BC的面积．\", \"type\": \"text\"}]}, {\"type\": \"answer\", \"blocks\": [{\"text\": \"(1)①见解析；②-3；(2)\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{7}{2}\"}]}, {\"type\": \"solution\", \"blocks\": [{\"text\": \"(1)①∵A(3, −1), A′(2, 2)，∴△ABC向左平移1个单位长度，向上平移3个单位长度得到△A′B′C′，如图，△A′B′C′即为所求． ②：△ABC上任一点D(x, y)平移后对应的点为D′(x + a, y + b)，所以 x −1 = x + a, y + 3 = y + b, ∴a = −1, b = 3, ∴ab = −3.故答案为：−3． (2)△B′BC的面积为\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"3 \\\\times 3 - \\\\frac{1}{2} \\\\times 2 \\\\times 3 - \\\\frac{1}{2} \\\\times 2 \\\\times 1 - \\\\frac{1}{2} \\\\times 1 \\\\times 3 = 9 - 3 - 1 - \\\\frac{3}{2} = \\\\frac{7}{2}\"}, {\"text\": \".\", \"type\": \"text\"}]}], \"source\": {\"bbox\": [30.22, 380.32, 418.58, 462.34], \"file\": \"人教版/七下/2026江苏七年级期末数学试卷+答案695.pdf\", \"regions\": [{\"bbox\": [30.22, 380.32, 418.58, 462.34], \"page\": 3}], \"section\": \"generic\", \"page_end\": 3, \"page_start\": 3, \"section_label\": \"\", \"answer_regions\": [{\"bbox\": [43.99800109863281, 231.12158203125, 486.0638427734375, 338.1268615722656], \"page\": 6}], \"answer_original\": \"(1)①见解析；②−3；(2)\\\\frac{7}{2}\", \"solution_original\": \"(1)①∵A(3, −1), A′(2, 2)，∴△ABC向左平移1个单位长度，向上平移3个单位 长度得到△A′B′C′，如图，△A′B′C′即为所求． ②：△ABC上任一点D(x, y)平移后对应的点为D′(x + a, y + b)，所以 x −1 = x + a, y + 3 = y + b, ∴a = −1, b = 3, ∴ab = −3.故答案为：−3． (2)△B′BC的面积为3 × 3 − \\\\frac{1}{2} × 2 × 3 − \\\\frac{1}{2} × 2 × 1 − \\\\frac{1}{2} × 1 × 3 = 9 −3 −1 − \\\\frac{3}{2} = \\\\frac{7}{2}.\"}, \"version\": 1, \"grade_level\": \"SENIOR\", \"number_label\": \"15\", \"subject_code\": \"math\", \"question_type\": \"comprehensive\", \"content_sha256\": \"39845afbda74088309536ede9392a4ad90b9083bbb0cabad8942357695df691b\"}","body_html":"<div>如图，在平面直角坐标系中，已知 △ ABC的三个顶点坐标分别是 A(3, −1), B(1, −4), C(3, −3)，将△ABC平移后得到△A′B′C′，其中点 A, B, C的对应点分别为A′, B′, C′．已知点A′的坐标是A′(2, 2)．<img class=\"question-figure\" src=\"http://tikupdfpng.mtwlkj.net:5129/qbank/math/2026/09/2b06312b-7e3f-494e-ae3e-21d10207fcf7/0db60c21-9484-43c8-8268-e05d5d6549de/828f6adb3fb3.png\" style=\"width:128.6px!important;max-width:100%;height:auto\" alt=\"\" title=\"点击查看原图\"><br>（1）①画出△A′B′C′； ②△ABC上任一点D(x, y)平移后对应的点为D′(x + a, y + b)，则ab =____ ；<br>（2）求△B′BC的面积．</div>","answer_html":"(1)①见解析；②-3；(2)\\(\\frac{7}{2}\\)","ways_html":"(1)①∵A(3, −1), A′(2, 2)，∴△ABC向左平移1个单位长度，向上平移3个单位长度得到△A′B′C′，如图，△A′B′C′即为所求． ②：△ABC上任一点D(x, y)平移后对应的点为D′(x + a, y + b)，所以 x −1 = x + a, y + 3 = y + b, ∴a = −1, b = 3, ∴ab = −3.故答案为：−3． (2)△B′BC的面积为\\(3 \\times 3 - \\frac{1}{2} \\times 2 \\times 3 - \\frac{1}{2} \\times 2 \\times 1 - \\frac{1}{2} \\times 1 \\times 3 = 9 - 3 - 1 - \\frac{3}{2} = \\frac{7}{2}\\).","explain_html":null,"difficulty":2,"score":null,"year":null,"parse_status":"PARSED","qa_status":"QUARANTINED","retry_count":0,"final_score":0.913,"error_message":"QA failed: COHERENCE,ANSWER","published_question_id":null,"content_sha256":"39845afbda74088309536ede9392a4ad90b9083bbb0cabad8942357695df691b","created_at":"2026-09-25T07:37:52","updated_at":"2026-09-25T07:49:10","answer_missing":false,"formula_fallback":false,"_number":"15","_section":"generic","_section_label":""},"assets":[{"id":4895,"asset_uuid":"d55c1149b48d404fa00dbe69ad07f42e","job_id":3,"import_question_id":22440,"asset_type":"embedded_image","source_file":"2026江苏七年级期末数学试卷+答案695.pdf","source_page":3,"source_bbox":"[30.22, 462.83, 126.68, 553.17]","sha256":"828f6adb3fb393378faf94a6c29fa773245544610105cb0f4d49addaf67bc619","local_path":"/www/wwwroot/tikupdfpng.mtwlkj.net_5129/qbank/math/2026/09/2b06312b-7e3f-494e-ae3e-21d10207fcf7/0db60c21-9484-43c8-8268-e05d5d6549de/828f6adb3fb3.png","asset_key":"math/2026/09/2b06312b-7e3f-494e-ae3e-21d10207fcf7/0db60c21-9484-43c8-8268-e05d5d6549de/828f6adb3fb3.png","public_url":"http://tikupdfpng.mtwlkj.net:5129/qbank/math/2026/09/2b06312b-7e3f-494e-ae3e-21d10207fcf7/0db60c21-9484-43c8-8268-e05d5d6549de/828f6adb3fb3.png","width":520,"height":430,"qa_status":"PENDING","created_at":"2026-09-25T07:40:08","safe_public_url":"http://tikupdfpng.mtwlkj.net:5129/qbank/math/2026/09/2b06312b-7e3f-494e-ae3e-21d10207fcf7/0db60c21-9484-43c8-8268-e05d5d6549de/828f6adb3fb3.png"}],"knowledge":[],"audits":[{"id":444851,"import_question_id":22440,"audit_type":"STRUCTURE","auditor_name":"structure-rule","status":"PASS","score":0.9,"input_snapshot":"如图，在平面直角坐标系中，已知 △ ABC的三个顶点坐标分别是 A(3, −1), B(1, −4), C(3, −3)，将△ABC平移后得到△A′B′C′，其中点 A, B, C的对应点分别为A′, B′, C′．已知点A′的坐标是A′(2, 2)． [图] (1) ①画出△A′B′C′； ②△ABC上任一点D(x, y)平移后对应的点为D′(x + a, y + b)，则ab =____ ； (2) 求△B′BC的面积．","result_json":"{\"ok\": true, \"rule\": \"clean-override\", \"issues\": [], \"confidence\": 0.9}","created_at":"2026-09-25T07:45:28"},{"id":444852,"import_question_id":22440,"audit_type":"FORMULA","auditor_name":"formula-auditor","status":"PASS","score":1.0,"input_snapshot":"\\frac{7}{2}; 3 \\times 3 - \\frac{1}{2} \\times 2 \\times 3 - \\frac{1}{2} \\times 2 \\times 1 - \\fr","result_json":"{\"risky\": 2, \"issues\": [], \"checked\": 2, \"invalid\": 0, \"visual_check\": \"delegated_to_question_vision\", \"invalid_details\": [], \"visually_verified\": 0}","created_at":"2026-09-25T07:45:28"},{"id":444853,"import_question_id":22440,"audit_type":"KNOWLEDGE","auditor_name":"knowledge-auditor","status":"WARN","score":0.0,"input_snapshot":"如图，在平面直角坐标系中，已知 △ ABC的三个顶点坐标分别是 A(3, −1), B(1, −4), C(3, −3)，将△ABC平移后得到△A′B′C′，其中点 A, B, C的对应点分别为A′, B′, C′．已知点A′的坐标是A′(2, 2)． [图] (1) ①画出△A′B′C′； ②△ABC上任一点D(x, y)平移后对应的点为D′(x + a, y + b)，则ab =____ ； (2) 求△B′BC的面积．","result_json":"{\"picks\": [], \"tree_code\": \"KT_SENIOR_MATH_STANDARD\", \"candidates\": 0, \"classifier_version\": null}","created_at":"2026-09-25T07:45:28"},{"id":444857,"import_question_id":22440,"audit_type":"VISION","auditor_name":"vision-auditor","status":"PASS","score":0.95,"input_snapshot":"题干：\n如图，在平面直角坐标系中，已知 △ ABC的三个顶点坐标分别是 A(3, −1), B(1, −4), C(3, −3)，将△ABC平移后得到△A′B′C′，其中点 A, B, C的对应点分别为A′, B′, C′．已知点A′的坐标是A′(2, 2)． （1）①画出△A′B′C′； ②△ABC上任一点D(x, y)平移后对应的点为D′(x + a, y + b)，则ab =____ ； （2）求△B′BC的面积．\n答案：\n(1)①见解析；②-3；(2)\\(\\frac{7}{2}\\)\n解析：\n(1)①∵A(3, −1), A′(2, 2)，∴△ABC向左平移1个单位长度，向上平移3个单位长度得到△A′B′C′，如图，△A′B′C′即为所求． ②：△ABC上任一点D(x, y)平移后对应的点为D′(x + a, y + b)，所以 x −1 = x + a, y + 3 = y + b, ∴a = −1, b = 3, ∴ab = −3.故答案为：−3． (2)△B′BC的面积为\\(3 \\times 3 - \\frac{1}{2} \\times 2 \\times 3 - \\frac{1}{2} \\times 2 \\times 1 - \\frac{1}{2} \\times 1 \\times 3 = 9 - 3 - 1 - \\frac{3}{2} = \\frac{7}{2}\\).","result_json":"{\"ok\": true, \"hard\": false, \"issues\": [], \"confidence\": 0.95, \"suggestions\": [\"建议在题干中明确说明平移方向（如‘向左平移1个单位，向上平移3个单位’），避免学生误判平移向量。\", \"建议在解析中补充对△B′BC面积计算方法的几何解释（如割补法原理），增强理解。\"]}","created_at":"2026-09-25T07:45:51"},{"id":444858,"import_question_id":22440,"audit_type":"SOLVER","auditor_name":"blind-solver","status":"PASS","score":1.0,"input_snapshot":"如图，在平面直角坐标系中，已知 △ ABC的三个顶点坐标分别是 A(3, −1), B(1, −4), C(3, −3)，将△ABC平移后得到△A′B′C′，其中点 A, B, C的对应点分别为A′, B′, C′．已知点A′的坐标是A′(2, 2)． [图] (1) ①画出△A′B′C′； ②△ABC上任一点D(x, y)平移后对应的点为D′(x + a, y + b)，则ab =____ ； (2) 求△B′BC的面积．","result_json":"{\"basis\": \"由A(3,-1)平移到A'(2,2)得平移向量为(-1,3)，故a=-1,b=3，ab=-3\", \"answer\": \"-3\", \"confidence\": 1.0, \"handout_chars\": 0}","created_at":"2026-09-25T07:46:26"},{"id":444859,"import_question_id":22440,"audit_type":"CRITIC","auditor_name":"critic","status":"PASS","score":1.0,"input_snapshot":"来源答案:(1)①见解析；②-3；(2)\\(\\frac{7}{2}\\) || 盲解:-3","result_json":"{\"ok\": true, \"issues\": [], \"merged\": true, \"confidence\": 1.0}","created_at":"2026-09-25T07:46:26"},{"id":444860,"import_question_id":22440,"audit_type":"SYMBOLIC","auditor_name":"symbolic-auditor","status":"PASS","score":0.9,"input_snapshot":"答案:(1)①见解析；②-3；(2)\\(\\frac{7}{2}\\) || 盲解:-3","result_json":"{\"checks\": [], \"detail\": {\"reason\": \"盲解与来源答案互相包含\"}, \"status\": \"PASS\", \"applicable\": true}","created_at":"2026-09-25T07:46:26"},{"id":444861,"import_question_id":22440,"audit_type":"COHERENCE","auditor_name":"coherence-auditor","status":"FAIL","score":0.8,"input_snapshot":"答案:(1)①见解析；②-3；(2)\\(\\frac{7}{2}\\) || 解析:(1)①∵A(3, −1), A′(2, 2)，∴△ABC向左平移1个单位长度，向上平移3个单位长度得到△A′B′C′，如图，△A′B′C′即为所求． ②：△ABC上任一点D(x, y)平移后对应的点为D′(x + a, y + b)，所以 x −1 = x + a, y + 3 = y + b, ∴a = −1, b = 3, ∴ab = −3.故答案为：−3． (2)△B′BC的面积为\\(3 ","result_json":"{\"ok\": false, \"which\": [], \"issues\": [\"3 × 3 - 1/2 × 2 × 3 - 1/2 × 2 × 1 - 1/2 × 1 × 3 = 7/2: 计算过程错误。正确计算应为：9 - 3 - 1 - 1.5 = 3.5，即 7/2。虽然最终结果正确，但解析中写的是 9 - 3 - 1 - 3/2 = 7/2，这一步是正确的，但前面的几何意义解释不够清晰，且没有明确说明如何得到这些数值（底和高）。\"], \"confidence\": 0.8, \"equation_checks\": [{\"valid\": false, \"reason\": \"计算过程错误。正确计算应为：9 - 3 - 1 - 1.5 = 3.5，即 7/2。虽然最终结果正确，但解析中写的是 9 - 3 - 1 - 3/2 = 7/2，这一步是正确的，但前面的几何意义解释不够清晰，且没有明确说明如何得到这些数值（底和高）。\", \"equation\": \"3 × 3 - 1/2 × 2 × 3 - 1/2 × 2 × 1 - 1/2 × 1 × 3 = 7/2\"}]}","created_at":"2026-09-25T07:46:40"},{"id":444862,"import_question_id":22440,"audit_type":"QUALITY_GATE","auditor_name":"quality-gate","status":"FAIL","score":0.913,"input_snapshot":"","result_json":"{\"warnings\": [], \"graded_gate\": true, \"hard_failed\": [\"COHERENCE\"]}","created_at":"2026-09-25T07:46:40"},{"id":444863,"import_question_id":22440,"audit_type":"SOLVER","auditor_name":"blind-solver","status":"PASS","score":1.0,"input_snapshot":"如图，在平面直角坐标系中，已知 △ ABC的三个顶点坐标分别是 A(3, −1), B(1, −4), C(3, −3)，将△ABC平移后得到△A′B′C′，其中点 A, B, C的对应点分别为A′, B′, C′．已知点A′的坐标是A′(2, 2)． [图] (1) ①画出△A′B′C′； ②△ABC上任一点D(x, y)平移后对应的点为D′(x + a, y + b)，则ab =____ ； (2) 求△B′BC的面积．","result_json":"{\"basis\": \"由A(3,-1)平移到A'(2,2)得平移向量(a,b)=(-1,3)，故ab=-3。B'坐标为(0,-1)，利用三点坐标面积公式计算△B'BC面积为7/2。\", \"answer\": \"-3; \\\\frac{7}{2}\", \"confidence\": 1.0, \"handout_chars\": 0}","created_at":"2026-09-25T07:48:15"},{"id":444864,"import_question_id":22440,"audit_type":"CRITIC","auditor_name":"critic","status":"PASS","score":1.0,"input_snapshot":"来源答案:-3;\\(\\frac{7}{2}\\) || 盲解:-3; \\frac{7}{2}","result_json":"{\"ok\": true, \"issues\": [], \"merged\": true, \"confidence\": 1.0}","created_at":"2026-09-25T07:48:15"},{"id":444865,"import_question_id":22440,"audit_type":"COHERENCE","auditor_name":"coherence-auditor","status":"FAIL","score":0.9,"input_snapshot":"答案:-3;\\(\\frac{7}{2}\\) || 解析:(1) ① 由 A(3,−1) 平移到 A′(2,2)，得平移向量为 (−1,3)，即向左平移1个单位，向上平移3个单位。据此画出△A′B′C′。 ② 平移规则为 D(x,y)→D′(x+a,y+b)，对比 A 与 A′ 坐标得 a=−1，b=3，故 ab=−3。 (2) 由平移得 B′(0,−1)。利用割补法：以矩形包围△B′BC，其面积为\\(3\\times 3=9\\)；减去周围三个直角三角形面","result_json":"{\"ok\": false, \"which\": [\"solution\"], \"issues\": [\"解析中割补法三个直角三角形面积计算错误，导致最终△B′BC面积结果错误。\", \"S₁=½×2×3=3: 直角三角形 S₁ 的两直角边应为 B(1,-4) 到 B'(0,-1) 的水平距离 1 和垂直距离 3，面积应为 ½×1×3=1.5，而非 ½×2×3=3。\", \"S₂=½×2×1=1: 直角三角形 S₂ 的两直角边应为 B(1,-4) 到 C(3,-4) 的水平距离 2 和 C 到 B' 的垂直距离 3（或类似组合），但根据割补法实际应为 ½×2×3=3 或其他，此处 ½×2×1=1 错误。\", \"S₃=½×1×3=1.5: 直角三角形 S₃ 的两直角边应为 C(3,-4) 到 C'(3,-1) 的垂直距离 3 和 C' 到 B' 的水平距离 3，面积应为 ½×3×3=4.5，而非 ½×1×3=1.5。\", \"S△B′BC=9−3−1−1.5=3.5=7/2: 由于 S₁、S₂、S₃ 计算均错，导致最终面积结果错误。正确面积应为 3（见下方正确推导）。\"], \"confidence\": 0.9, \"equation_checks\": [{\"valid\": false, \"reason\": \"直角三角形 S₁ 的两直角边应为 B(1,-4) 到 B'(0,-1) 的水平距离 1 和垂直距离 3，面积应为 ½×1×3=1.5，而非 ½×2×3=3。\", \"equation\": \"S₁=½×2×3=3\"}, {\"valid\": false, \"reason\": \"直角三角形 S₂ 的两直角边应为 B(1,-4) 到 C(3,-4) 的水平距离 2 和 C 到 B' 的垂直距离 3（或类似组合），但根据割补法实际应为 ½×2×3=3 或其他，此处 ½×2×1=1 错误。\", \"equation\": \"S₂=½×2×1=1\"}, {\"valid\": false, \"reason\": \"直角三角形 S₃ 的两直角边应为 C(3,-4) 到 C'(3,-1) 的垂直距离 3 和 C' 到 B' 的水平距离 3，面积应为 ½×3×3=4.5，而非 ½×1×3=1.5。\", \"equation\": \"S₃=½×1×3=1.5\"}, {\"valid\": false, \"reason\": \"由于 S₁、S₂、S₃ 计算均错，导致最终面积结果错误。正确面积应为 3（见下方正确推导）。\", \"equation\": \"S△B′BC=9−3−1−1.5=3.5=7/2\"}]}","created_at":"2026-09-25T07:49:10"},{"id":444866,"import_question_id":22440,"audit_type":"SYMBOLIC","auditor_name":"symbolic-auditor","status":"WARN","score":0.55,"input_snapshot":"答案:-3;\\(\\frac{7}{2}\\) || 盲解:-3; \\frac{7}{2}","result_json":"{\"checks\": [], \"detail\": {\"reason\": \"盲解与来源答案存在字面差异，需 Critic 复核\"}, \"status\": \"WARN\", \"applicable\": true}","created_at":"2026-09-25T07:49:10"},{"id":444867,"import_question_id":22440,"audit_type":"AUTO_REPAIR_ATTEMPT","auditor_name":"answer-repair","status":"FAIL","score":1.0,"input_snapshot":null,"result_json":"{\"critic\": true, \"symbolic\": true, \"coherence\": false, \"consensus\": \"two_solvers+symbolic+critic+coherence\", \"confidence\": 1.0, \"verifier_base\": \"http://127.0.0.1:8091/v1\", \"verifier_confidence\": 1.0, \"critic_explicit_pass\": true, \"symbolic_literal_warn\": false, \"symbolic_explicit_pass\": false, \"coherence_explicit_pass\": false}","created_at":"2026-09-25T07:49:10"}]}