The junior high school formulas mainly used in high school solid geometry

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Then press these to expand to the face.

The others are parallel, perpendicular.

Just get the vector ones right.

There are books on it... You're a tutor.

Congruence, similar knowledge should be used, and some properties of the circle will also be often used, such as the diameter of the circle is a right angle to the circumferential angle, the corresponding line segment of the parallel line cut is proportional, and the properties of the parallelogram will be commonly used.

All the formulas for high school solid geometry are as follows:

1. Cube a-side length s=6a2; v=a3。

2. Cuboid a-long; b-wide; c-high; s=2（ab+ac+bc）；v=abc。

3. Cylindrical r-bottom radius; h-high; c—circumference of the bottom surface; S bottom - bottom area; S side — side area. S surface - surface area is noisy, c = 2 r, s bottom = r2, s side = ch, s table = ch + 2s bottom, v = s bottom h = r2h.

4. Hollow cylinder r-outer radius; r-inner circle radius; h-high; v=πh（r2-r2）。

5. Straight cone r-bottom radius; h-high v = r2h3.

6. Round table r - upper bottom radius r - lower bottom radius h - high, v= h(r2+rr+r2) 3.

7. Prismatic S-bottom area; h-high; v=sh。

8. Pyramid S-bottom area h-high; v=sh/3。

9. The upper and lower bottom areas of the prism S1 and S2 are h-high; v=h[s1+s2+（s1s1）1/2]/3。

10. Pseudo-cylinder S1-upper and bottom area; s2-lower bottom area; s0-mid-sectional area; h-high; v=h（s1+s2+4s0）/6。

11. a ball r-radius; d - diameter, v = 4 3 r3 = d2 6.

12. Ball lack H-ball lack high; r-ball radius; a-radius of the ball missing the bottom, v= h(3a2+h2) 6= h2(3r-h) 3, a2=h(2r-h).

13. The ball table R1 and the lifting and cleaning die R2 - the upper and lower bottom radius of the ball table; H-high, v = h[3(r12+r22)+h2]6.

14. a toroidal r-torus radius; d-Ring diameter; r-torus cross-sectional radius; D-Ring cross-section forward diameter v=2 2rr2= 2dd2 4.

15. Barrel body D-barrel belly diameter; d-Diameter of the bottom of the barrel; H - barrel height, v = h (2d2 + d2) 12 (the bus bar is arc-shaped, the center of the circle is the center of the barrel) v = h(2d2 + dd + 3d2 4) 15 (the bus bar is parabolic).

Cube incision sphere r=a 2;Outside ball r=a 3 2;Ridge catchr = a 2 2;

Regular tetrahedral incision ball r=a 6 12;Outside ball r=a 6 4;Ridge catches r=a 2 4;

Quadrangular pyramid inscribed ball r=a( 6- 2) 4;Outside ball r=a 2 2;Ridge catches r=a 2;

1) Incut Ball:

The inscribed ball is tangent to all sides of the cube, so radius r=a 22) outward catch:

The vertices of the cube are on the outside ball, and the diagonal length of the cube is 2r, r=(root number 3)a 2

The edge catches the ball through the midpoint of the cube, and the face angle of the cube is 2r long, r = (root number 2) a 2

ab should be the diameter.

acb=90°

PO Bottom ABC

po⊥abpa=pb=6

ao=ob=4

po=2√5

Connect to the OMOM PB

om=1/2pb=3

The magnitude of the angle formed by the radius OC and the bus PB is equal to 60°, i.e. MOC = 60°

Cosine theorem.

cos60°=(om +oc -cm) (2*om*oc) to get cm= 13

Over m as mn ab in n

ab//po

Mn = 5, the angle formed by the heterogeneous straight line MC and Po, i.e., cmn

cos cmn=mn cm= 5 13= 65 13 The angle formed by the heterogeneous straight line MC and PO = arccos 65 13 If you agree with my answer, please click "Accept as satisfactory answer" in the lower left corner, and wish you progress in learning!

What does ab and oc mean when they are the radius of the base circle, and what is the center of the bottom circle?

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