![](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 bottom surface of four pyramids P-ABCD is square, and the bottom surface of PB ⊥ ABCD proves that no matter how the height of four pyramids PB changes, the surface PAD and the surface PCD cannot be vertical
Let Pb = h, the side length of the square is a, make a vertical line from C to PD in the face PCD, and intersect PD with E
∵ Pb ⊥ bottom ABCD
∴PB⊥CD、PB⊥AD
∵PB⊥CD、CD⊥BC
Therefore, DC ⊥ PC and ⊥ PCD are right angles
Similarly, ∠ pad is a right angle,
∵CD=AD,PD=PD,∠PAD=∠PCD=90
⊥ PD can be proved by △ PCD and △ pad congruence, so ∠ CEA is the dihedral angle of face pad and face PCD
Let CE = AE = B, ∵ s △ PCD = DC * PC / 2 = PD * B / 2
∴b=a*sqrt(h^2+a^2)/sqrt(h^2+2a^2)
As shown in the figure, PA and Pb tangent to o at point a and B. m are a point on circle O. through M, EF is tangent to circle O, intersection PA and Pb at e and F, and PA = 12cm As shown in the figure, PA and Pb tangent circle O at point a and B.M are a point on circle O, and EF is tangent to circle O through M, intersect PA and Pb at e and F, and PA = 12cm
PB=PA=12
According to the tangent property, EA = em, FB = FM
So the perimeter of the triangle PEF = PE + pf + EF = PE + pf + Em + FM = (PE + EA) + (PF + FB) = PA + Pb = 24
As shown in the figure, PA and Pb tangent o to points a and B respectively, and the angle P is equal to 58 degrees. C is a point on the circle O. find the angle C
Connect OA and ob
∵ PA and Pb cut ⊙ o at points a and B respectively,
∴OA⊥PA、OB⊥PB,
∵∠P=58°,
∴∠AOB=122°,
∴∠C=61°．
If Pb = 4cm, PC = 16cm, find the length of PA The relationship between a circle and a straight line is an exercise
|PA | squared = | Pb | PC |
The length of PA is 8
If Pb = 4cm. PC = 16cm (1, find the length of PA (2. Prove: PA ^ 2 = Pb × PC)
Pc-pb = BC = 12 (BC is diameter)
OB=6
PO=10
OA=6
PA=8
2.PA^2=8*8=64
PB×PC=4*16=64
PA^2=PB×PC
Given: ∠ AOB and two points c and D, find a point P such that PC = PD, and the distance from point P to both sides of ∠ AOB is equal (requirement: draw with ruler and gauge, keep the trace of drawing, write the method, no proof is required)
As shown in the figure:
Methods: (1) draw an arc with o as the center of the circle and any length as the radius, and intersect OA and ob at two points respectively;
(2) Take the two intersection points as the center of the circle, and take the length greater than half of the distance between the two intersection points as the radius;
(3) Take o as the end point, draw a ray at the intersection point inside the corner;
(4) Connect CD, C and D as the center of the circle, greater than 1
The length of 2CD is the radius of arc, which intersects two points respectively;
(5) Draw a straight line through two intersections;
(6) This line intersects with the ray drawn earlier at point P,
The point P is the point to be calculated
Given: ∠ AOB and two points c and D, find a point P such that PC = PD, and the distance from point P to both sides of ∠ AOB is equal (requirement: draw with ruler and gauge, keep the trace of drawing, write the method, no proof is required)
As shown in the figure:
Methods: (1) draw an arc with o as the center of the circle and any length as the radius, and intersect OA and ob at two points respectively;
(2) Take the two intersection points as the center of the circle, and take the length greater than half of the distance between the two intersection points as the radius;
(3) Take o as the end point, draw a ray at the intersection point inside the corner;
(4) Connect CD, C and D as the center of the circle, greater than 1
The length of 2CD is the radius of arc, which intersects two points respectively;
(5) Draw a straight line through two intersections;
(6) This line intersects with the ray drawn earlier at point P,
The point P is the point to be calculated
Draw a picture: find a point P so that PC = PD, and the distance between P and AOB is equal
The point P is the point
Known: as shown in the figure, ∠ AOB = 30 ° P is a point on the bisector of ∠ AOB, PC ‖ OA, intersection ob with point C, PD ⊥ OA, the perpendicular foot is D, if PC = 4, find the length of PD
Passing point P is PE ⊥ OB,
∵PC∥OA，
∴∠CPO=∠POD，
∵ OP is the bisector of AOB,
∴∠COP=∠DOP，
∴∠COP=∠CPO，
∵∠AOB=30°，
∴∠PCE=30°，
∵PC=4，
∴PE=2，
The length of PD is 2
As shown in the figure, ∠ AOB = 30 ° OC bisection ∠ AOB, P is any point on OC, PD ‖ OA intersects ob in D, PE ⊥ OA in E. if od = 4cm, find the length of PE
Pass P as PF ⊥ OB to F,
∵ AOB = 30 ° OC bisection ∠ AOB,
∴∠AOC=∠BOC=15°，
∵PD∥OA，
∴∠DPO=∠AOP=15°，
∴∠BOC=∠DPO，
∴PD=OD=4cm，
∵∠AOB=30°，PD∥OA，
∴∠BDP=30°，
In RT △ PDF, PF = 1
2PD=2cm，
∵ OC is angle bisector, PE ⊥ OA, PF ⊥ ob,
∴PE=PF，
∴PE=PF=2cm．
RELATED INFORMATIONS
- 1. As shown in the figure, ∠ AOB = 30 ° OC bisection ∠ AOB, P is a point on OC, PD ‖ OA intersects ob at point D, PE ⊥ OA at e, OD = 4cm, then PE ⊥ OA is at point E=______ .
- 2. Known: as shown in the figure, find a point P on the straight line Mn, so that the distance between the point P and the two sides of ∠ AOB is equal (the method is required to be written, the drawing trace shall be retained, and the conclusion shall be written)
- 3. In the plane rectangular coordinate system, the image of function y = - 3 / 4x + 6 intersects X axis and Y axis at point a and B respectively, line BC and X axis intersect at C, and point C is the midpoint of line OA Find out whether there is a point m on the straight line AB so that △ BCM is an isosceles triangle. If there is, find out the coordinates of point M
- 4. As shown in the figure, the ray OA, ob, OC, OD have common endpoint o, and ∠ AOB = 90 °, COD = 90 °, AOD = 5 4 ﹤ AOC. calculate the degree of ﹤ BOC
- 5. As shown in the figure, OA, OB and OC are the radii of circle O. if ∠ AOB = 2 ∠ BOC, please judge whether ∠ ACB = 2 ∠ BAC is tenable. Why?
- 6. The right angle vertex P of the triangle ruler slides on the ray OM, and the two right angle sides intersect OA and ob at point CD respectively Conjecture the quantitative relationship between PC and PD
- 7. As shown in the figure, the side of the cone is cut along OA, and the fan-shaped OAB is expanded into a plane figure (1) What is the relationship between the length of the sector arc AB and the length of the circle at the bottom of the cone? What is the position relationship between point a and point B on the side of the cone? (2) If the angle ∠ AOB = 90 °, what is the relationship between the radius r of the cone bottom circle and the radius r of fan-shaped OAB (i.e. OA or OB)? (3) If point a moves on the side of the cone and then returns to its original position, how should the shortest distance of point a be designed? If R 2 = 0.5, ∠ AOB = 90 °, find the shortest distance of point a
- 8. As shown in the figure, it is known that ∠ AOB = 30 ° and the point P is inside ∠ AOB, Op = 6. If there is a moving point m on OA and a moving point n on ob, then the minimum perimeter of △ PMN is______ .
- 9. In the triangle ABC, if AC = radical 2, BC = radical 7, ab = 3, then cosa =?
- 10. Known: as shown in the figure, in the right triangle ABC, the angle c = 90 degrees, the point O is the center of the circle, the length of OA is the radius of the circle, and AC, AB intersect with point D.E, and the angle CBD = angle a.bd is the tangent line of circle O. find: if ad: Ao = 8:5, BC = 2, find the length of BD
- 11. It is known that: as shown in the figure, P is the interior point of the square ABCD, ∠ pad = ∠ PDA = 15 °. Verification: △ PBC is an equilateral triangle
- 12. Ray OA, ob, OC, OD have a common endpoint o, and OA is perpendicular to ob, OC is perpendicular to OD, ∠ AOD = 5 / 4 ∠ AOC, calculate the degree of ∠ BOC ∠ AOD = 5 / 4 ∠ AOC! No, ∠ AOD = 5 / 4 ∠ BOC
- 13. As shown in Fig
- 14. It is proved that the coordinates of the point P which makes the vector AP * vector BP get the minimum value is [emergency] given that the point a (2,0), B (4,0), and the moving point P move in the parabola y ^ 2 = - 4x, it is proved that the coordinate of the point P which makes the vector AP * vector BP get the minimum value is the coordinate origin. Please give the detailed process,
- 15. Mathematical problem: introduce two tangent lines PA, Pb from the moving point P to the circle x ^ 2 + y ^ 2 = 1 1, introduce two tangent PA, Pb from the moving point P to the circle x ^ 2 + y ^ 2 = 1, and the tangent points are a, B, and ∠ APB = 60 degrees, then the trajectory equation of the moving point P is________ 2. Let P (x, y) be a moving point on the circle x ^ 2 + (Y-1) ^ 2 = 1. If the inequality x + y + C > 0 holds, then the value range of C_____________ 3. We know that the equation of the line where the edge ab of the equilateral △ ABC is located is x √ 3 + y = 0, and the coordinates of point C are (1, √ 3) 4. It is known that the circle M: x ^ 2 + (Y-2) ^ 2 = 1, q is the moving point on the x-axis, QA and QB tangent the circle m to two points a and B respectively (1) (liberal arts) if | ab | = 4 √ 2 / 3, find the equation of line MQ (Science) prove that the straight line AB always crosses a fixed point (2) Find the locus equation of midpoint P of moving string ab It's better to analyze it
- 16. As shown in the figure, lead the tangent PA, Pb of the circle through a point p outside ⊙ o with radius of 6cm, connect Po, intersect ⊙ o at F, make tangent of ⊙ o through F, intersect PA and Pb at D and e respectively, if Po = 10cm, ﹤ APB = 40 ° Find: (1) the circumference of △ ped; (2) the degree of ∠ doe
- 17. Pass through the plane α of the triangle ABC and a point p outside the plane α of △ ABC, make Po ⊥ α, and connect PA, Pb, PC with the perpendicular foot of O 2. If PA = Pb = PC, then o is △ ABC__ heart
- 18. As shown in the graph, P is a point in the equilateral △ ABC, PA = 6, Pb = 8, PC = 10, if the point P 'is a point outside △ ABC, but △ p'ab is congruent with △ PAC The distance between point P and point P 'and the degree of ∠ APB
- 19. As shown in the figure, BC is the diameter of ⊙ o, P is a point on ⊙ o, a is the midpoint of arc BC, ad ⊥ BC, perpendicular foot is D, Pb intersects AD and AC respectively at points e, F, AE and be? Why? It stops after 20 minutes
- 20. It is known that P is a point in the plane of triangle ABC. If vector CB = γ vector PA + vector Pb, γ belongs to R, then point P must be in the plane A. Inside the triangle ABC B. On the line where the AC side is located C. On the line where AB side is located D. On the line where the BC side is located Why?