High School Math Geometry Questions About Outside Catch

1. Find the volume of the regular tetrahedral outer receiving ball with an edge length of 1.

Analysis: The regular tetrahedron is A-BCD

Through A as the Ao Plane BCD Intersection BCD in O, O is the center of the BCD.

The connection bobo = 3 2*2 3= 3 3ab = 1 by the Pythagorean theorem ao 2=1-1 3=2 3==>ao= 6 3 Let the radius of the outside ball be r, and the center of the ball o' must be on ao.

Connect o'b==>o'b=o'a=r

then o'b 2-oo' 2=ob 2

r 2-( 6 3-r) 2 = 1 3 = = > r = 6 4 volume v = 4 * r 3 3 = 6 8

2. The ratio of the surface area of the ball to its inner cube is.

Analysis: Let the radius of the sphere be r, and the edge length of the inner cube should be a

The length of the diagonal of the inscribed cube is the diameter of the ball2r

a^2+a^2+a^2=4r^2

a^2=4r^2/3

The surface area of the cube is.

s=6a^2=8r^2

The surface area of the ball is s1=4 r2.

The ratio of the surface area of the ball to its inscribed cube is 2

3. It is known that the vertices of a cube are all on the surface of the sphere, and if its edge length is 4cm, then the surface area of the sphere is.

From question 2, we know a 2=4r 2 3==>r 2=3 4a 2a=4, and the surface area of the ball is: s=48

or the ratio of the surface area of the ball to its inscribed cube is 2

s=96*π/2=48π

4. It is known that the vertices of a cube are on the spherical surface, if the volume of the sphere is, then the surface area of the cube is.

Analysis: Ball v=4 3 r 3=

Ball r=3 4, by question 2 a 2=4r 2 3=3 4 The surface area of the cube s=6a=6*3 4=9 2

1. The tetrahedron is A-BCD, and the high line AO intersection plane BCD is in O, O is the center of BCD, at this time, BOA is a right triangle, Bo = (root number 3 2) * (2 3) = (root number 3) 3;

ab =1 Therefore, from the Pythagorean theorem ao 2=1-1 3=2 3 ao= (root number 6) 3;

The radius of the outside ball is r = (2 3) * ao = (2 * root number 6) 9 volume v = 4 * r 3 3

2, yes 2

The length of the diagonal of the cube is the diameter of the ball2r

And because a 2 + a 2 + a 2 = 4r 2

a^2=4r^2/3

The surface area of the cube is.

s=6a^2=8r^2

The surface area of the ball is 4 r 2.

3. By s 16*6 = 2

The surface area of the ball is: s=48

4. v = 4 3 r 3 = so: r = 3 2, the surface area of the cube s = 6a = 6 * 4 3r = 18

The high school math outside ball universal formula is sphere volume 4 3*(d 2)3.

Analysis: The spatial diagonal of the cuboid is the diameter of the outer ball, so first find the spatial diagonal of the cuboid a b c. Know the diameter, then divide by 2 to get the radius. Then the volume is obtained according to the volume formula of the ball.

Basic Introduction:

The center of the polygon inscribed sphere is the intersection of all the bihedral planes of the polygon.

The position of the polygon circumscribed cherry blossom ball ball o can be determined in one of the following ways:

1. The point o is the intersection of two straight lines that are connected to the center of the circle by the non-parallel plane of the polyhedron and perpendicular to the non-parallel plane.

2. Point o is the intersection of the three planes that are represented by the midpoint of the non-parallel edges of the polyhedron and perpendicular to the three planes of these edges.

3, point o is the intersection of the straight line perpendicular to the plane of the circle and the plane perpendicular to the midpoint of the edge that is not parallel to the surface of the circumscribed circle center of a surface.

For rotating bodies and polyhedra, the outside ball has different definitions of perturbation dispersion, which is broadly understood as the ball surrounds the frontal geometry, and the vertices and arcs of the slow geometry are on this ball.

The outboard ball of the cube and cuboid is the intersection of their spatial diagonals.

The outer ball of the round table is a circle that passes through the upper and lower circles and the arc distance from the center of the circle to the two circles is equal.

High school math outside ball solving skills are as follows:

1) Grasp the key characteristics of "joint" and "cut".

a) Outside catch.

The key feature of the outside ball is the outside "catch". Therefore, the distance from each "connection" point to the center of the sphere is equal and equal to the radius, which should be used well when solving the problem, whether it is constructing a figure or calculating the old age.

b) Incision ball.

The key feature of the in-cut ball is the in-cut ball. Therefore, the distance from each "tangent" point to the center of the sphere is equal and equal to the radius, and the line connecting with the center of the sphere is perpendicular to the facet, which should be used when solving the problem, whether it is to construct the figure or calculate.

2) Capture the "center position" feature.

In this kind of problem, the center of the combination is often located in some special positions (such as the center of the circle and the center coincide) due to some properties of the combination (such as symmetry), so in many cases determining the center position is very important for solving the problem. The general method is:

a) Determining the central location is generally a critical first step in solving a problem.

When it is an outside ball or only one inward ball, the center of the combination is the center of the ball; When there is more than one inscribed ball, and the two are tangent to each other, the center position can be determined according to the symmetry, the center perpendicular line of the inside surface of the outside ball, and other characteristics.

b) Constructing the geometry, which is usually the critical second step in solving the problem (and then just calculate the fundamentals and substitute the formula to solve it).

Based on the position of the center and the center of the sphere (when it does not coincide with the center), combined with the outer blind junction or the inner tangent point, a geometry can be easily used to assist in the calculation – the final goal is mostly a right triangle. This is the key to solving this kind of problem.

Because , PA surface ABCD, it can be known that PA AC.

And because , abcd is a square , we can know that ac = ab + ad

pc² = pa² +ac²

So, pc is the diameter of the ball. i.e. pc = 2r.

It can be proved that the triangle PBC, PAC, PDC is a right-angled triangle with a common hypotenuse PC, and the distance from the midpoint of the PC O, O to A, B, and D is equal to half of the hypotenuse PC. i.e. oa=ob=od=op=oc=1 2pc

So the diameter of the outside ball is 2r=pc

An inscribed sphere is a sphere surrounded by geometry that is equal to its radius from its center to each face. (Not all geometry has an outside ball).

Intuitive perspective.

Dissect it and look at it.

An outboard ball is a ball that surrounds the geometry and the vertices of the geometry are on the ball. (Not all geometry has an outside ball).

Intuitive perspective.

Dissect it and look at it.

In fact, learning is the power of thinking.

The inner diagonal of 1 cube is the circumscribed circle.

2. The center of the side of the cube is an inscribed circle.

An inscribed sphere is a ball with an equal distance from the center of the sphere to the geometry (all faces) at a point in space, and an external sphere is a ball with an equal distance from the center of the sphere to each (vertex) of the space geometry at a point in space.

Because the hall spring is two pairs perpendicular, so s pab=pa*pb 2, and so on.

Let Pa=a, Pb=B, and Pc=C

Soshan shouted with ab=3, ac=2, bc=12

Find: a= 2 2, b=3 2, c=2 The surface area of the outside ball = 4 r 2= (a 2 + b 2 + c 2) = 53 2 cm 2