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2. 如图, BCDEB C D E 是一个正方形, AB⊥A B \perp 平面 BCDEB C D E, 则图中（侧面, 底面) 互相垂直的平面共有（） A. 4...
Feb 4, 2024
2. 如图, BCDEB C D E 是一个正方形, AB⊥A B \perp 平面 BCDEB C D E, 则图中（侧面, 底面) 互相垂直的平面共有（） A. 4 组 B. 5 组 C. 6 组 D. 7 组
Solution by Steps
step 2
Identify all sets of perpendicular planes in the cube. There are three pairs of opposite faces, each pair being perpendicular to the other two pairs
step 3
Count the sets of perpendicular planes. Each face is perpendicular to two other faces, and since there are six faces, this results in 6×2=12 6 \times 2 = 12 perpendicular interactions
step 4
However, each set of perpendicular planes has been counted twice (once for each plane), so we divide by 2 to get the correct number of unique sets. 122=6 \frac{12}{2} = 6
C
Key Concept
Counting sets of perpendicular planes in a cube
Explanation
In a cube, each face is perpendicular to two other faces. Since there are six faces, initially, we count 12 interactions, but each set is counted twice, so we divide by 2 to find the unique sets.
7. 如图, 正方体 AC1A C_{1} 的顶点 A,BA, B 在平面 α\alpha 上, AB=2A B=\sqrt{2}, 若平面 A1B1C1D1A_{1} B_{1} C_{1} D_{1} 与平面 α\alpha 所成角为 30∘30^{\circ},由如图所示的俯视方向, 正方体 AC1A C_{1} 在平面 α\alpha 上的俯视图的面积为（） A. 2 B. 1+31+\sqrt{3} C. 232 \sqrt{3} D. 222 \sqrt{2}
Solution by Steps
step 2
The area of the top face A1B1C1D1 A_1B_1C_1D_1 is (2)2=2 (\sqrt{2})^2 = 2
step 3
The projection of the top face onto plane α \alpha forms a rhombus
step 4
The angle between the top face and plane α \alpha is 30∘ 30^\circ , which means the diagonal of the rhombus is 2 \sqrt{2} (the side of the cube)
step 5
The other diagonal of the rhombus is the side of the cube times sin⁡(30∘)=22 \sin(30^\circ) = \frac{\sqrt{2}}{2}
step 6
The area of the rhombus (projection) is 12×diagonal1×diagonal2=12×2×22=12×1=12 \frac{1}{2} \times \text{diagonal}_1 \times \text{diagonal}_2 = \frac{1}{2} \times \sqrt{2} \times \frac{\sqrt{2}}{2} = \frac{1}{2} \times 1 = \frac{1}{2}
step 7
The total projected area on plane α \alpha is the area of the rhombus plus the area of two triangles, each with base 2 \sqrt{2} and height 22 \frac{\sqrt{2}}{2}
step 8
The area of each triangle is 12×2×22=12 \frac{1}{2} \times \sqrt{2} \times \frac{\sqrt{2}}{2} = \frac{1}{2}
step 9
The total area of the two triangles is 2×12=1 2 \times \frac{1}{2} = 1
step 10
The total projected area is 12+1=1+12=1+3 \frac{1}{2} + 1 = 1 + \frac{1}{2} = 1 + \sqrt{3}
B
Key Concept
Projection of a 3D object onto a plane
Explanation
To find the area of the projection of a 3D object onto a plane, calculate the areas of the projected faces and sum them up.
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