Knowing that the edge length of the cube is 1, find the surface area and volume of its inscribed and

Inscribed ball: Since it is a cube.

, then the edge length of the cube is the diameter of the inscribed sphere, so the radius of the inscribed sphere is a 2. Circumscribed sphere: Since it is an inscribed sphere of a cube, and the eight vertices are all on the surface of the inscribed sphere, the diagonal of the body of the cube is the diameter of the inscribed sphere, that is, a which is 3 times the root number, so the radius of the inscribed sphere is a that is 3 times the root number of the half.

Spheres tangent to each edge: take an edge on the upper surface of the cube as the object of study; Since the sphere is tangent to the edges of the cube, the distance from the upper surface to the center of the sphere is half the length of the edges, i.e., a2;The surface intercepted by the top surface is the inscribed circle of the upper surface.

denoted as truncated circle), then the radius of the circle is a 2, it can be seen that the distance from the truncated circle center to the center of the sphere is a 2, and the distance from each edge in the upper surface to the center of the circle is a 2 (i.e., the distance from the tangent point to the truncated center of the circle is a 2); A right triangle can be formed from the center of the sphere (denoted O), the truncated center of the circle (denoted A), and the tangent point (denoted B).

Among them, OA and AB are two right-angled edges, and OB is the hypotenuse.

oa = ab = a 2, then ob = a of 2 times the root number of the half, that is, the radius of the ball tangent to each edge is a of 2 times the root number of the second part.

The formula for calculating the surface area of the ball is as follows: surface area of the ball = 4 r 2, r is the radius of the ball

The formula for calculating the volume of the ball is: v ball = (4 3) r 3, r is the radius of the ball and the radius of the inscribed ball is 0 5, and the radius of the outside ball is the root number of two.

The body diagonal of the cube a1c 2 = 4 2 + 4 2 + 4 2 = 48, a1c = 4 3, the radius of the outer hand ant catching the ball a1o = 2 3, the surface area of the ball = 4 (2 3) 2 = 48

The volume of the outside ball is buried = 4 3 (2 3) 3 = 32 3

The diameter of the inscribed sphere = the edge length of the cube = 4, the diameter of the semi-noisy mass is 2, and the surface area = 4 2 2 = 16

The radius of the inscribed ball is a 2

Outside ball radius (cube diagonal) 2

The tangent to each edge should also be half the diagonal.

The diameter of the inscribed ball is 1, then the radius is 1 2 v = (4 3) 1 2) cubic = 6

The distance from the center of the inscribed ball of the cube O to the surface of each discard of the cube is equal to the radius, 2r=2, that is, the radius of the sphere r=1, and the rent of the inscribed ball is 4

So the answer is: 4 ;

According to the title, the diagonal of the cube (i.e., the diameter of the outside ball) is 3, so the radius of the outside ball is 3 2So the surface area of the outside ball according to the formula s=4 r2=3Volume v=4 3* r3=3 2

Let the tetrahedron be ABCD and the center of BCD (the center of the triangle, the outer center, the inner center, and the vertical center in one) be E, then the center of the ball O is on AE, and r=ao=3*od=3ae 4.

Since BE=2(3)3, AE=2(6)3.

r=ao=3ae/4=(√6)/2。

v=(4πr^3)/3=(√6)π。

32a，o2e=36

A, connect Bo2, in RT Bo2E, Bo2=63, connect A1 and Bo2 to O3, by RT AO3 rt Bo1O2, O3O2=O3O1, O3A=O3B, the same can be proved O3C=O3D=O3A, the distance from O3 to the other two sides is also equal to O3O1, O3 is the center of the ball outside and inside the tetrahedron, by Bo1O3 Bo2E, O1O3=612

a, r outside = 64

a, outside = 32 a

r within = 612

a, within s = 16 a

will be a tetrahedron.

Make up a cube to judge the shirt buried, and the edge of the tetrahedron is the face of the square to dig the ant's body to face the diagonal.

If the edge length of the tetrahedron is 2, then the edge length of the cube is 2

Its in-cut ball. is the inside of the cube radius r = 2 2 s1 = 4 r 2 = 2 The outside ball is the outside of the cube, the diameter = the diagonal of the cube = 6 radius r = 6 2 s2 = 4 r 2 = 6