If the bottom of p-abcd is a parallelogram, the bottom of PA ⊥ is ABCD, the point E is on the side edge PC, and PE = 13pc, then VP − bdevp − ABCD=______ .
Let the area of parallelogram ABCD be s, the height of pyramid p-abcd be h, ∵ PA ⊥ plane ABCD, ∵ PA = H.S △ abd = s △ BCD = 12s, ∵ PE = 13pc, the height of pyramid e-bcd be 23h, ∵ vp-bde = vp-abcd-vp-abd-ve-bcd = 13sh-13 × 12sh-13 × 23h × 12s = (13 − 16 − 19
RELATED INFORMATIONS
- 1. PA is perpendicular to the plane of the square ABCD with side length a, PA = a, then the size of the dihedral angle b-pc-d is It needs a rough process
- 2. Through the vertex a of the square ABCD, make the line PA ⊥ plane ABC, and PA = Pb, then the degree of the sharp dihedral angle formed by plane ABC and plane PCD is? Sorry, wrong number. It's not pa = Pb, it's PA = ab
- 3. Given that PA ⊥ square ABCD, if AB = PA, then the dihedral angle of plane PAB and plane PCD is
- 4. If the line AP is perpendicular to the plane ABCD through the vertex a of the square ABCD, and AP = AB, then what is the dihedral angle between the plane ABP and the plane CDP? (Figure)
- 5. If the vertex a of the square ABCD passes through the AP vertical plane ABCD and AP = AB, then the degree of the dihedral angle between the plane ABP and the plane DCP is Draw a picture
- 6. If the vertex a of the square ABCD passes through the line AP ⊥ plane ABCD, and AP = AB, then the degree of the dihedral angle formed by the plane ABP and the plane CDP is If you do well, you can pay me a picture, because I don't know how to draw a plane CDP
- 7. Through the vertex a of square ABCD, do PA ⊥ plane ABCD, PA = AB = 1, find the dihedral angle formed by plane APB and plane cpd
- 8. Given that ABCD is a rectangle, ab = a, ad = B, PA ⊥ plane ABC, PA = C, then the distance from point P to BC and BD is How to find two distances?
- 9. Given that ABCD is a rectangle, ab = a, ad = B, PA ⊥ plane ABC, PA = C, then the distance from P to BC is? And the distance from P to BD is?
- 10. Given that PA is perpendicular to the plane of rectangle ABCD, and ab = a, ad = B, PA = 2c, find the distance from the midpoint Q of PA to the straight line BD
- 11. The bottom of p-abcd is a parallelogram, the bottom of PA ⊥ is ABCD, e is a point on PA, and the section of PC ∥ is BDE Find the volume ratio of the two parts of the pyramid p-abcd divided by the section BDE
- 12. The bottom surface of the pyramid p-abcd is a square with side length 1, and the side edge PA is perpendicular to the bottom surface ABCD. Moreover, PA = 2. If e is the midpoint of PA, we prove the plane BDE
- 13. P is a point outside the parallelogram ABCD, and Q is the midpoint of PA
- 14. P is a point out of the plane of the parallelogram ABCD. Find a point E on PC to make the PA ‖ face be, and give the proof
- 15. P is the point out of the plane of the parallelogram ABCD, and Q is the midpoint of PA
- 16. P is a point outside the parallelogram ABCD, and Q is the midpoint of PA
- 17. Let p be a point in the square ABCD, and the distances from point P to vertex ABC are 1, 2 and 3 respectively
- 18. P is a point inside the square ABCD. The distances from point P to vertex a, B and C are 1, 2 and 3 respectively. Find the side length of the square Using trigonometric function solution, there are two results under the solution. Why should 5-2 √ 2 be omitted?
- 19. A mathematical problem: let p be a point inside the square ABCD, and the distances from P to vertex a, B and C are 1, 2 and 3, respectively
- 20. In the cube abcd-a1b1c1d1, M is the midpoint of CC1. If point P is on the plane of abb1a1 and satisfies ∠ pdb1 = ∠ mdb1, then the trajectory of point P is () A. Circle B. ellipse C. hyperbola D. parabola