1. The radius of circle O is 6cm, the chord AB = 10cm, the chord CD = 8cm, and ab is perpendicular to CD and P

1. The radius of circle O is 6cm, the chord AB = 10cm, the chord CD = 8cm, and ab is perpendicular to CD and P

In a right triangle OEA, OE ⊥ ab ⊥ e, of ⊥ CD in F, then oepf is rectangular. In a right triangle OEA, OE ⊥ 6 ⊥ 5 ⊥ 11. In the right triangle OFC, it is easy to obtain the diagonal of the rectangular oepf: OP ⊥ ab ⊥ e, of ⊥ CD ⊥ f ⊥ so the length of OP ⊥ is ⊥ 3

It is known that the diameter of circle O is 10cm, the chord AB is 6cm, and P is a point on ab. if the length of OP is an integer, how many points satisfy the condition?

OP ∈ [4,5], so there are three points satisfying the condition, which are the midpoint of point a, point B and point ab

As shown in the figure, the diameter of ⊙ o is 10cm, chord AB is 8cm, and P is a point on chord ab. if the length of OP is an integer, then the point P satisfying the condition is () A. 2 B. Three C. Four D. Five

As shown in the figure, connect OA and make OD ⊥ AB to D through o,
⊙ the diameter of O is 10cm, and the chord AB is 8cm,
When op ⊥ AB, Op has a minimum value,
Then ad = 1
2AB=4cm，
Od obtained from Pythagorean theorem=
OA2−AD2 =
52−42=3cm，
When op ⊥ AB, the minimum value of OP is 3,
When OP and OA coincide, the maximum P is 5,
In AD, there are 3, 4, 5 integer points in the middle of P,
There are four, five, two integer points between BD,
So p has five integer points on ab
Therefore, D

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

As shown in the figure, given that AB is the diameter of ⊙ o, chord CD ⊥ AB at point P, CD = 10cm, AP: Pb = 1:5, then the radius of ⊙ o is () A. 6cm B. 3 5m C. 8cm D. 5 Three

Let AP = x, then Pb = 5x, then the radius of ⊙ o is 1
2（x+5x）=3x
∵ chord CD ⊥ AB at point P, CD = 10 cm
∴PC=PD=1
2CD=1
2×10=5cm
According to the intersecting string theorem, CP · PD = AP · Pb
That is, 5 × 5 = x · 5x
The solution is X=
5 or X=-
5 (omitted)
So the radius of ⊙ o is 3x = 3
5cm，
Therefore, B

AB is the diameter of circle O, the chord CD is perpendicular to P, CD = 10cm, AP: Pb = 1:5, the radius of circle O is? Question supplement:

Let the radius be 3x (3x) ^ 2 = (2x) ^ 2 + 5 ^ 2 to get x = 1 and radius to 3

As shown in the figure, if the radius of ⊙ o is 5cm, P is the point outside ⊙ o, Po = 8cm, ∠ P = 30 °, then ab=______ cm．

As shown in the figure: make OD ⊥ AB in D and connect ob,
Because ∠ P = 30 °
So od = 1
2PO=1
2×8=4cm
In the right triangle ODB, BD=
OB2−OD2 =
52−42=3cm
According to the vertical diameter theorem, if BD = ad, then AB = 2bd = 2 × 3 = 6cm

As shown in the figure, it is known that AB is the chord of ⊙ o, and P is a point on ab. if AB = 10cm, Pb = 4cm, Op = 5cm, then the radius of ⊙ o is equal to______ cm．

Extend OP to both sides,
If OC = xcm, then CP = (x + 5) cm, PD = (X-5) cm,
According to the intersecting string theorem, AP? BP = CP? DP, i.e. (10-4) × 4 = (x + 5) (X-5),
The solution is x2 = 49, x = 7 or x = - 7 (negative value is rounded off),
Then the radius of ⊙ o is equal to 7cm

If the radius of ⊙ o is 5cm and the chord center distance of chord AB is 3cm, then the length of chord AB is______ cm．

As shown in the figure, OD ⊥ AB is at point D, according to the vertical diameter theorem,
Point D is the midpoint of AB, OD = 3cm, OA = 5cm,
AD=1
2AB=
OA2−OD2=4，
∴AB=8cm．