Gaokao Solid Geometry Proofs and Calculations Flashcards

In the pyramid P-ABCD, the base ABCD is a rectangle. E is the midpoint of the side PD. Let plane α be the plane passing through points A, E, and C. What is the relationship between line PB and plane α?

Answer: PB is parallel to plane α

Let O be the intersection of the diagonals AC and BD of the rectangle. Connect EO. In triangle PBD, E is the midpoint of PD and O is the midpoint of BD, so EO is the midsegment. Therefore, EO is parallel to PB. Since EO lies within plane AEC (plane α) and PB does not, PB is parallel to plane α.

In a cube ABCDA'B'C'D' with side length 'a', what is the sine of the angle between the line A'B and the plane BDD'B'?

Answer: 1/2

Set up a coordinate system with D at the origin (0,0,0). Then A'=(a,0,a), B=(a,a,0). The vector for line A'B is v = B - A' = (0, a, -a). The plane BDD'B' is the plane x=y, so its normal vector is n = (1, -1, 0). The sine of the angle θ between a line and a plane is given by sin(θ) = |v · n| / (|v| * |n|) = |(0)(1) + (a)(-1) + (-a)(0)| / (sqrt(a²+a²) * sqrt(1²+1²)) = a / (a√2 * √2) = a / (2a) = 1/2.

In a triangular pyramid P-ABC, the base ABC is an equilateral triangle with a side length of 2. If PA is perpendicular to the base plane ABC and PA = 3, what is the volume of the pyramid P-ABC?

Answer: √3

The volume of a pyramid is V = (1/3) * Base Area * Height. The height is given as PA = 3. The area of the equilateral triangle base is (√3/4) * side² = (√3/4) * 2² = √3. Therefore, the volume V = (1/3) * √3 * 3 = √3.

In a regular tetrahedron with an edge length of 3, what is the distance from any vertex to the opposite face?

Answer: √6

The distance from a vertex to the opposite face is the height (H) of the regular tetrahedron. The formula for the height is H = a√(2/3) = a√6/3, where 'a' is the edge length. For a=3, the height is H = 3√6/3 = √6.

In the right triangular prism ABC-A₁B₁C₁, ∠BCA = 90°, and AC = BC = CC₁ = 1. What is the cosine of the dihedral angle B₁-AC₁-B?

Answer: 2√2/3

Set up a coordinate system with C at (0,0,0). Then A=(1,0,0), B=(0,1,0), C₁=(0,0,1), B₁=(0,1,1). The two planes are Plane(B,A,C₁) and Plane(B₁,A,C₁). A normal vector to Plane(B,A,C₁) is n₁ = vector(AC₁) × vector(AB) = (1,0,1) × (-1,1,0) = (-1,-1,1). A normal vector to Plane(B₁,A,C₁) is n₂ = vector(AC₁) × vector(AB₁) = (1,0,1) × (-1,1,1) = (-1,-2,1). The cosine of the angle θ between the planes is |n₁·n₂| / (|n₁||n₂|) = |1+2+1| / (√3 * √6) = 4/√18 = 4/(3√2) = 2√2/3.

A rectangular box has side lengths of 2, 3, and 6. What is the surface area of the sphere that circumscribes this box?

Answer: 49π

The diameter of the circumscribing sphere is equal to the space diagonal of the rectangular box. The square of the diagonal is d² = l² + w² + h² = 2² + 3² + 6² = 4 + 9 + 36 = 49. So, the diameter d = 7, and the radius R = 3.5. The surface area of the sphere is A = 4πR² = 4π(3.5)² = 4π(12.25) = 49π.