![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
The square ABCD is folded into a straight dihedral angle along the AC side, and the out of plane line AB and CD form an angle of? Online, etc., urgent, urgent
ninety
Given that AB is the diameter of circle O, ab = 10, the distance between point a and line CD is 3, and the distance between point B and line CD is 6, then the position relationship between line CD and circle O is ()
Because the center O is the midpoint of ab
Then the distance from the center O to CD = (3 + 6) / 2 = 4.5
As shown in the figure, AB is the diameter of ⊙ o, ad is the chord of ⊙ o, the tangent passing through point B intersects the extension line of ad at point C. If ad = DC, find the degree of ∠ abd
∵ BC is the tangent line of ⊙ o, ∵ ab ⊥ BC, ∵ ABC = 90 °, ∵ AB is the diameter of ⊙ o, ∵ ADB = 90 °, ∵ BD ⊥ AC, ad = CD, ∵ ABC is isosceles right triangle, ∵ BD bisects ⊥ ABC, ∵ abd = 45 °
Given the circle X & # 178; + Y & # 178; = 9. A point p outside the circle is tangent PA, Pb (a, B are tangent points). When the point P moves on the straight line 2x-y + 10 = 0, the minimum area of the quadrilateral paob is
The quadrilateral paob is two symmetrical right triangles, OA = 3 is the radius of the circle PA & # 178; = Po & # 178; - OA & # 178; = Po & # 178; - 9 is the area of the quadrilateral paob = 2 △ Pao area = 2 * 1 / 2 * PA * OA = 3PA. To minimize the area, PA is required to be the minimum, that is, Po is required to be the minimum. When Po is perpendicular to the line, Po is the minimum distance from the origin (Center) to the line, Po = i0-0 + 10I / √ (1 + 4) = 2 √ 5 Pa & # 178; =If 20-9 = 11 PA = √ 11, the minimum area of the quadrilateral paob is 3 √ 11
I want to ask why the area of paob is 3PA? I don't understand
Area of quadrilateral paob = 2 △ Pao area = 2 * 1 / 2 * PA * OA = 3PA (because OA = 3 is given by the title)
I don't understand why "the distance from the origin (Center) to the straight line Po = i0-0 + 10I / √ (1 + 4) = 2 √ 5
It is known that circle C: (x-1) ∧ 2 + (Y-2) ∧ 2 = 2, passing through point P (2, - 1) to be tangent of circle C, and the tangent points are a and B
(1) The equation of finding straight line PA, Pb
(2) Find the tangent length of the circle C passing through point P
(1) Let the slope of tangent be K, then the equation is y + 1 = K (X-2)
That is kx-y - (2k + 1) = 0
The radius of the center of the circle (1,2) is √ 2, and the distance from the center of the circle to the tangent is equal to the radius, which is substituted into the formula of the distance from the point to the straight line
Yes √ 2 = | K-2 - (2k + 1) | / √ (k ^ 2 + 1)
The solution is k = 3 ± √ 2
Write your own equations
(2) The radius of tangent perpendicular to the tangent point is 2, | PC | = distance between two points, formula = √ [(2-1) ^ 2 + (- 1-2) ^ 2] = √ 10
According to Pythagorean theorem, the tangent length L = 2 √ 2
Given the circle C: (x-1) (Y-2) = 2, the coordinates of point P are (2, - 1), the tangent of circle C is made through P, and the tangent points are a and B. the equations of straight lines PA and PA are obtained
The second question is to find the tangent length of the circle passing through point P, and the third question is to find the equation of line ab
There are two points AB, which is a and which is B, otherwise PA line cannot be determined
A (- 2,0) B (2,0) C is a moving point on the circle x ^ 2 + y ^ 2 = 1
I'm wrong, it's B (0, 2)
The area of the triangle depends on the distance from the line AB to the point
As can be seen from the graph, the closest point to the line segment is (- radical 2 / 2, radical 2 / 2). In this way, the distance between the point and the straight line is calculated
Let a (6,0), B (0,6) and C be the points on the ellipse X & # 178 / 20 + Y & # 178 / 5 = 1, and find out the minimum area of △ ABC. With the solution of trigonometric function, the graph can be obtained
Look at the picture
The tangent point PA, Pb, a, B of the leading circle x ^ 2 + y ^ 2 = 1 passing through the point P (0,4) is used to solve the linear equation passing through the tangent point a, B_________
Draw a circle with P as the center and PA as the radius
Then PA ^ 2 = OP ^ 2-oa ^ 2 = 16-1 = 15
So the circle is
x^2+(y-4)^2=15
The equation of subtracting circle
We get y = 1 / 4
The bisector BM of AB = AC, DB = DC, ∠ ABC intersects ad at m, passes through ⊙ o of B and m, intersects BC at point G, intersects AB at point F, and FB is exactly the diameter of ⊙ o
(1) Ad is the bisector of ∠ BAC
(2) Ad is the tangent of ⊙ o
Certification:
（1）
∵AB=AC,DB=DC,AD=AD
∴⊿ABD≌⊿ACD（SSS）
∴∠BAD=∠CAD
That is, ad is the bisector of ∠ BAC
（2）
∵ AB = AC, ad bisection ∠ bac
⊥ ad ⊥ BC [isosceles triangle three lines in one]
Connect OM
∵OB=OM
∴∠OMB=∠OBM
∵∠ OBM = ∠ CBM [BM bisection ∠ ABC]
∴∠OMB=∠MBC
∴OM//BC
∴OM⊥AD
The tangent of circle O is ad
RELATED INFORMATIONS
- 1. In the circle O, AB is the diameter, the chord CD intersects the circle O at points c, D, and is parallel to ab. the distance between AB and CD is 5, CD = 24, and ab is calculated
- 2. As shown in the figure, AB is perpendicular to plane BCD, BC is perpendicular to CD, ab = BC, and the angle between AD and plane BCD is 30 degrees Finding the angle between AD and plane ABC Picture in
- 3. (1 / 2) in the pyramid p-abcd, the bottom surface ABCD is rectangular, the vertical plane ABCD of PA, PA = ad = 1, ab = 2, e and F are the midpoint of AB and PD respectively（
- 4. Given that the coordinates of point P (x, y) satisfy the condition x + y ≤ 4Y ≥ XX ≥ 1, and point O is the origin of coordinates, then the minimum value of | Po | is equal to___ , the maximum value is equal to___ ．
- 5. As shown in the figure, AB and CD are the two chords of circle O. points E and F are the midpoint of AB and CD, connecting EF, angle AEF = angle CFE, and proving AB = CD
- 6. It is known that the plane angle of dihedral angle α - ab - β is an acute angle, C is a point in plane α (it is not on edge AB), D is the projection of point C on plane β, and E is the projection of point C on edge ab
- 7. In the dark house, there is a plane mirror on the table. Behind the mirror is a white wall. When you shine a flashlight on the mirror and the wall, the wall is very bright because the wall has a very good effect on the light______ The mirror is dark because it makes the light shine______ ．
- 8. The intersection of two planes forms the plane angle How many parts of the space are the three planes that do not coincide with each other? Some say 8, some say 7, and 8
- 9. Plane angle vertical plane method of dihedral angle What is the vertical plane method in the common method of making plane angle of dihedral angle?
- 10. How to find the plane angle of dihedral angle?
- 11. (1) Find the angle formed by the straight lines PD and AB on different planes (2) prove PA perpendicular to CD (3) find the size of dihedral angle p-ab-d In the pyramid p-abcd, the side PDC is an equilateral triangle with side length of 2 and perpendicular to the bottom ABCD, the bottom ABCD is a diamond with area of 2.3, the angle ADC is an acute angle of the diamond, and M is the midpoint of Pb
- 12. The square ABCD is folded into a dihedral angle with a plane angle of 120 ° along the diagonal BD to calculate the size of the dihedral angle b-ac-d The square ABCD is folded into a dihedral angle with a plane angle of 120 ° along the diagonal BD Finding the size of dihedral angle b-ac-d
- 13. In a triangular pyramid a-bcd, plane abd ⊥ plane DBC, ∠ ABC = ∠ DBC = 120 ° AB = BC = BD, the tangent of plane angle of dihedral angle a-bd-c is obtained Urgent ah, homework, ask experts to help
- 14. The waist length of isosceles triangle is 10cm, the area is 3cm2 under 25 heel trigonometric function
- 15. If a triangle with three sides BC = 3, AC = 4, ab = 5 is folded into △ ABC 'along the longest side AB, then the length of CC' is () A. 125B. 512C. 245D. 524
- 16. It is known that the area of triangle ABC is s, the acute angle between plane ABC and plane a is Q, the projection of triangle ABC on plane a is triangle a'b'c ', and the area of triangle a'b'c is s'. This paper proves that s' = s * cosq (please tell me the proof idea,
- 17. Three questions about solid geometry in senior high school, 1. In cube AC1, e is the point on edge CC1, and a = C1E / EC, (1) if plane bed1 ⊥ plane bdd1b1, then a=_____ (2) if bed1 ⊥ ab1c, then a=_____ . 2. (1) in the diamond ABCD, a = 60 ° AB = 4, after folding it into a straight dihedral angle along BD, AC = 4=_____ The tangent of dihedral angle a-cd-b is equal to_____ . (2) In rectangle ABCD, ab = 3, ad = 4, it is folded into a straight dihedral angle along BD, AC=_____ . 3. It is known that ABCD is a rectangle, PA ⊥ plane ABCD, M is the midpoint of PC, PA = ad (1) Verification: plane mAb ⊥ plane PCD (2) Finding the size of dihedral angle m-ab-c
- 18. Seeking high school solid geometry problem It's a little difficult to find some intersection lines in high school solid geometry. It's also a proof of solid geometry
- 19. When is it necessary to make auxiliary lines and when is it not necessary to make auxiliary lines?
- 20. Straight line and plane vertical exercises The triangle ABC has a point p outside the plane α, through which PQ is made, the vertical plane α is made, and the perpendicular foot is O, connecting PA, Pb and PC (1) If PA = Pb = PC, then o is triangular ABC___ Heart (2) If PA vertical Pb, Pb vertical PC, PC vertical PA, then o is triangular ABC___ Heart (3) If the distances from P to three sides AB, BC and Ca are equal, then o is a triangle ABC___ Heart (4) If PA = Pb = PC, ∠ C = 90 °, then o is ab edge___ . (5) If PA = Pb = PC, ab = AC, then point O is at___ on-line. There is no picture in the original question. Yes, please send me the answer and the answer