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In the triangular vertebrae with o as the vertex, the intersection angle of the three edges passing through O is 30 degrees. Take two points AB on one edge, OA = 4cm, OB = 3cm. At the end point of AB, a rope is used to tightly wrap the side of the triangular pyramid (there is no friction between the rope and the side), and find the shortest rope length between AB and ab
The angle AOB is 30 * 3 = 90 degrees when the side of triangular prism is expanded
Because the line segment between two points is the shortest
Shortest rope length = AB = 5
Solid geometry The height of a cylinder is 20cm, the radius of its bottom is 5cm, and the volume of an inscribed cuboid is 800cm * 2. How to calculate the surface area of the cuboid
In this paper, we first point out the defects of the title. The volume unit is cm * 3V = s (base area) × height. Let the length, width and height of the cuboid be ABC respectively. We can see that the height C = (the height of the cylinder) 20cm ﹥ AB = 800 ﹤ 20 = 40 cm - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
A straight line is drawn through the midpoint of any two edges of the parallelepiped abcd-a1b1c1d1, among which the straight lines parallel to the plane dbb1d1 share () A. 4 B. 6 C. 8 D. 12
As shown in the figure, make a straight line through the midpoint of any two edges of the parallelepiped abcd-a1b1c1d1,
There are 12 straight lines parallel to the plane dbb1d1,
Therefore, D
A high school mathematics solid geometry problem In tetrahedral PABC, PA.PB.PC Two pairs of perpendicular, M is a point in plane ABC, and m to three faces PAB.PBC.PCA If the distance between M and vertex P is 2.3.6, what is the distance from m to vertex p? There needs to be an analysis~
If PABC is regarded as the four adjacent vertices of a rectangle, then MP = (2 * 2 + 3 * 3 + 6 * 6) is divided by 2 and then the solution is obtained
It is known that the Aa1 ⊥ plane ABC, ab = BC = Aa1 = Ca, P is a point on A1B (1) When a1p ⊥ Pb, AB is better than PC (2) When the dihedral angle p-ac-b is 60 degrees, the value of a 1p is greater than that of Pb Some symbols, like "I can't type..." ... tomorrow! I can't do it= [wrong number on the top, sorry sorry ~] It is known that the Aa1 ⊥ plane ABC, ab = BC = Aa1 = Ca, P is a point on A1B. (1) When a1p is higher than Pb, ab ⊥ PC (2) When the dihedral angle p-ac-b is 60 degrees, the value of a1p is higher than that of Pb.
(1) When = 1
Let PD ∥ A1A intersect AB with D and connect CD. From A1A ⊥ surface ABC, we know PD ⊥ surface ABC. When p is the midpoint of A1B, D is the midpoint of ab. ∵ ⊥ ABC is a regular triangle,
Given that the straight line y = x + 0.5 and parabola y ^ 2 = 4x and the fixed point E (1,2), m, n are two moving points on the parabola and satisfy EM vertical en, it is proved that the line Mn always passes through the fixed point No idea I want the process I don't understand
Mn is set to (m ^ 2 / 4, m) (n ^ 2 / 4, n) (M and N > 0)
According to the definition of vertical
(m^2/4-1)(n^2/4-1)+(m-2)(n-2)=0
We get (M + 2) (n + 2) + 16 = 0
Write a straight line in 2-point form
y-m=4/(m+n) *(x-m^2/4)
y=(4x+mn)/m+n
So over the fixed point (5,2)
Given the direction vector V = (1, root 3) of the line L crossing point (0, - 2, root 3) and ellipse C: X ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a is greater than B is greater than 0) and the eccentricity of ellipse is root 6 / 3 (1) Solving the equation of elliptic C (2) If the point d (3,0) is known, the point m n is the two points which do not coincide on the ellipse, and DM = KDN (with vector symbol), the value range of real number k is calculated First, I can do it, but second, I'm in trouble My idea is to set up the equation passing through the point D, set up the elliptic equation, eliminate y, and then according to the discriminant to a range of slope, and use Weida theorem to get the relationship X1 + x2 Then, according to the law of fixed score point, the relationship between X1 + kx2 is obtained. Finally, K is expressed with only slope However, it can not be expressed Let l be the root of the line L = 0, and then we can get the direction of the line So (2,0) is the right focus of the ellipse. C knows. If we know a according to the eccentricity, we can find B That's right
We know that the direction vector is v = (1, √ 3) of the line L through the point (0, - 2 √ 3) and the right focus of the ellipse C: x? 2 / a? + y? / b? = 1 (a > b > 0), and the eccentricity of the ellipse e = √ 6 / 3. (1) find the equation of ellipse C (2) if known
The straight line L: y = kx-4 has two different intersections m, N in parabola C: y ^ 2 = 8x
The simple solution to this problem is as follows: replace the straight line y = kx-4 into the parabola y ^ 2 = 8x to obtain (kx-4) ^ 2 = 8x, and then K ^ 2 * x ^ 2-8 (K + 1) x + 16 = 0 can be obtained by arranging the straight line y = kx-4 into the parabola y ^ 2 = 8x, and then K ^ 2 * x ^ 2-8 (K + 1) x + 16 = 0 can be obtained
Given that a (- 4,4), B (5,7), point E is the intersection point of AC and BD, e is in the first quadrant and the distance from Y axis is 1, the moving point P (x, y) still moves on one side of the rectangle BC (x is not equal to 0), and the value range of Y / X is obtained
If point B (- 3,2a-7), C (6,2a-4) and C (6,2a-4) in the rectangle ABCD, the point B (- 3,2a-7), C (6,2a-4) in the rectangle ABCD has 124ac | =
1. If the side length of a pyramid is equal to the side length of the bottom polygon, what shape can the pyramid be? 2. If the side length of a pyramid is equal to the side length of the bottom polygon, the pyramid may be a regular hexagonal pyramid; the answer is wrong If all the edges of a hexagonal pyramid are equal in length, then the bottom polygon is a regular hexagon. According to geometry, if a regular hexagon is taken as the base, the length of the side edge must be greater than that of the base If the side length of the pyramid is equal to the side length of the bottom polygon, what kind of figure can the pyramid be? I have a poor spatial imagination. I am a liberal arts student in senior three, The length of the Hexapod is longer than that of the base of the pyramid. In the regular hexagon, OA is the same as the side length ab. So the edge length is larger than the bottom side length If there is such a figure that the length of the side edge of the pyramid is equal to the side length of the bottom polygon, then as long as the distance from the center to each vertex is less than the side length of the positive polygon, then there will be such a figure. Then the graphics that meet the conditions are the regular three four five pyramids. The students are good. Is this my idea right?
A regular pyramid is an isosceles triangle whose sides are equal and congruent. It does not say that the sides and the sides of the base are equal. So you are talking about a special pyramid. There is no name. There is only a special regular pyramid in high school mathematics
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