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修改昨日录入的032052题面

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zhugemengtian 2024-01-22 11:43:57 +08:00
parent a822796f5b
commit 96b0891dae
1 changed files with 2 additions and 2 deletions
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题库0.3/Problems.json
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@ -693377,14 +693377,14 @@
},
"032052": {
"id": "032052",
"content": "如图, 在四棱锥$P-ABCD$中, 已知$PA \\perp$平面$ABCD$, 且四边形$ABCD$为直角梯形, $\\angle ABC=\\angle BAD=\\dfrac{\\pi}{2}$, $PA=AD=2$, $AB=BC=1$.\n\\begin{center}\n\\begin{tikzpicture}[>=latex]\n\\draw (0,0,0) node [above right] {$A$} coordinate (A);\n\\draw (0,0,1) node [left] {$B$} coordinate (B);\n\\draw (B) ++ (1,0,0) node [below] {$C$} coordinate (C);\n\\draw (2,0,0) node [right] {$D$} coordinate (D);\n\\draw (0,2,0) node [above] {$P$} coordinate (P);\n\\draw ($(B)!0.5!(P)$) node [left] {$Q$} coordinate (Q);\n\\draw (P)--(B) (P)--(C) (P)--(D) (B)--(C)--(D) (Q)--(C);\n\\draw [dashed] (B)--(A)--(D) (A)--(P); \n\\end{tikzpicture}\n\\end{center}\n(1) 求四棱锥$P-ABCD$的表面积;\\\\\n(2) 若$P, A, C, D$四点在同一球面上, 求该球的体积.",
"content": "如图, 在四棱锥$P-ABCD$中, 已知$PA \\perp$平面$ABCD$, 且四边形$ABCD$为直角梯形, $\\angle ABC=\\angle BAD=\\dfrac{\\pi}{2}$, $PA=AD=2$, $AB=BC=1$.\n\\begin{center}\n\\begin{tikzpicture}[>=latex]\n\\draw (0,0,0) node [above right] {$A$} coordinate (A);\n\\draw (0,0,1) node [left] {$B$} coordinate (B);\n\\draw (B) ++ (1,0,0) node [below] {$C$} coordinate (C);\n\\draw (2,0,0) node [right] {$D$} coordinate (D);\n\\draw (0,2,0) node [above] {$P$} coordinate (P);\n\\draw ($(B)!0.5!(P)$) node [left] {$Q$} coordinate (Q);\n\\draw (P)--(B) (P)--(C) (P)--(D) (B)--(C)--(D) (Q)--(C);\n\\draw [dashed] (B)--(A)--(D) (A)--(P); \n\\end{tikzpicture}\n\\end{center}\n(1) 求异面直线$PC$与$AB$所成角的大小; \\\\\n(2) 求四棱锥$P-ABCD$的表面积;\\\\\n(3) 若$P, A, C, D$四点在同一球面上, 求该球的体积.",
"objs": [],
"tags": [
"第六单元",
"2023届高三-第二轮复习讲义-09-立体几何综合"
],
"genre": "4em",
"ans": "(1) $\\dfrac 92+\\dfrac{\\sqrt{5}}2+\\sqrt{3}$; (2) $\\dfrac{8\\sqrt{2}}3\\pi$",
"ans": "(1);(2) $\\dfrac 92+\\dfrac{\\sqrt{5}}2+\\sqrt{3}$; (3) $\\dfrac{8\\sqrt{2}}3\\pi$",
"solution": "",
"duration": -1,
"usages": [],