The diameter of a circle is 10, and a chord AB is 8. P on AB, how many p points are there when OP is an integer?

The diameter of a circle is 10, and a chord AB is 8. P on AB, how many p points are there when OP is an integer?

Five

As shown in the figure, the diameter of ⊙ o is 10, the chord AB = 8, and P is a moving point on ab. find the range of Op

① When point P coincides with point a or point P, Op = r = 5;
② As shown in the figure:
∵OP⊥AB，
∴AP=PB=1
2AB=4，
In RT △ OPB, Op=
OB2−BP2=3．
In conclusion, the value range of OP is: 3 ≤ op ≤ 5

The radius of circle O is 6cm, chord AB = 10cm, chord CD = 8cm, and ab ⊥ CD is in P. find the length of Op

Do OE perpendicular to ab at point E and of perpendicular to CD at point F, then:
In the triangle ODF, of ^ 2 = 6 ^ 2-df ^ 2 = 36-16 = 20
In the triangle OEA, OE ^ 2 = 6 ^ 2-ae ^ 2 = 36-25 = 11
Easy to know: PE = of
In the triangle ope, Op ^ 2 = OE ^ 2 + PE ^ 2 = OE ^ 2 + of ^ 2 = 20 + 11 = 31
Therefore, the length of OP is root 31

If the diameter of ⊙ o is 10, chord AB = 8, and P is a moving point on chord AB, then the value range of OP length is () A. 3≤OP≤5 B. 4≤OP≤5 C. 4≤OP≤8 D. 8≤OP≤10

According to the shortest vertical line segment, OP is the shortest when op ⊥ ab
∵ diameter is 10, chord AB = 8
∴∠OPA=90°，OA=5，AP=4
The shortest OP is
OA2−AP2=3
∴3≤OP≤5
Therefore, a

As shown in the figure, ⊙ O's diameter AB bisects chord CD, CD = 10 cm, AP: Pb = 1:5

Connect Co, let the radius of the circle be r, ∵ diameter AB bisecting chord CD, ᙽ AB vertical CD (2 points) ∵ AP: Pb = 1:5, ᙽ if AP = k, Pb = 5K, then AB = AP + Pb = 6K, ᙽ OA = 3k, Po = oa-ap = 3k-k = 2K, ᙽ Po = 23oa = 23R (3) R2 = 52 + (23R) 2, R2 = 45

As shown in the figure, ⊙ O's diameter AB bisects chord CD, CD = 10 cm, AP: Pb = 1:5

Connect Co, let the radius of the circle be r, ∵ diameter AB bisecting chord CD, ᙽ AB vertical CD (2 points) ∵ AP: Pb = 1:5, ᙽ if AP = k, Pb = 5K, then AB = AP + Pb = 6K, ᙽ OA = 3k, Po = oa-ap = 3k-k = 2K, ᙽ Po = 23oa = 23R (3) R2 = 52 + (23R) 2, R2 = 45

In trapezoid ABCD, AB / / CD, angle d = 90, AB is the diameter of circle O, and ab = AD + BC. It is proved that CD is the tangent of circle o

Do OE / / AD and submit to E
So OE is the center line of trapezoidal ABCD, and OE is perpendicular to CD
So OE = 1 / 2 (AD + BC)
Because AB = AD + BC
So AB = 2oe
So OE is the radius of the circle o
Because CD passes through point E and OE is perpendicular to CD
So CD is tangent to circle o

It is known that in trapezoid ABCD, AB is parallel to CD, angle a = 90 degrees, BC is the diameter of circle O, BC = CD + ab

Then BCO is the diameter of the circle BCO,
Then Abe is the midpoint of the connecting line;
Because, OE ∥ AB,
Moreover, ab ⊥ ad,
Therefore, OE ⊥ ad;
Because OE = (CD + AB) / 2 = BC / 2,
That is: OE is the radius of the circle o,
Moreover, OE ⊥ ad,
So, ad is the tangent of circle o

As shown in the figure, if the quadrilateral ABCD is inscribed in ⊙ o, ad: BC = 1:2, ab = 35, PD = 40, then the tangent length of ⊙ o passing through point P is () A. 60 B. 40 Two C. 35 Two D. 50

As a tangent PE, according to the cutting line theorem, Pe2 = PD · PC = PA · Pb, so PAPC = pdpb, and △ pad and △ PBC have a common angle P, ∠ PDA = ∠ PBC, so △ pad ∽ PBC

As shown in the figure, in the quadrilateral ABCD, the midpoint of E, F, G and H is ab, BC, CD and Da respectively, Confirmation: eg and FH were equally divided

Proof: connect EF, FG, GH, he, AC
F is the midpoint of F, and ab is ab
The EF is the median line of △ ABC
∴EF‖AC,EF=1/2AC
Similarly, Hg is the median line of △ ACD
∴GH‖AC,HG=1/2AC
∴EF =HG ,EF ‖HG
The quadrilateral efgh is a parallelogram
The EG and FH are equally divided