It is known that in the circle O, if the chord AB / / CD, ab = 2 roots, 11 cm, CD = 4 roots, 5cm, and the distance between the chord AB and CD is 1cm, then the radius of circle O is?

It is known that in the circle O, if the chord AB / / CD, ab = 2 roots, 11 cm, CD = 4 roots, 5cm, and the distance between the chord AB and CD is 1cm, then the radius of circle O is?

If AB and CD are on the same side of circle 0, make the chord center distance of AB and CD, and the perpendicular feet are e and F. then let the distance OE from the center of circle O to CD be x, and the distance from circle O to ab be (1 + x)
Solve two right triangle OAE and OCF,
Solution x = 4, r = 6
AB and CD are on both sides of circle 0, and the value of X is meaningless

If AB is the diameter of circle O, the chord CD ⊥ AB is e, AE = 16, be = 4, then CD = ()

Because OE = 6, Pythagorean theorem CE = 8, and because CE = ed, CD = 16

As shown in the figure, ⊙ o diameter AB and chord CD intersect at point E, AE = 2, EB = 6, ∠ DEB = 30 ° to find the chord CD length

In RT △ OEF, ∠ DEB = 30 ° and ν of = 12oe = 1. In RT △ ODF, of = 1, OD = 4. According to the Pythagorean theorem, DF = OD2 is obtained

As shown in the figure, ⊙ o diameter AB and chord CD intersect at point E, AE = 2, EB = 6, ∠ DEB = 30 ° to find the chord CD length

In RT △ OEF, ∠ DEB = 30 ° and ν of = 12oe = 1. In RT △ ODF, of = 1, OD = 4. According to the Pythagorean theorem, DF = OD2 is obtained

As shown in the figure, the diameter of circle O AB and chord CD intersect at point E, AE = 2cm, EB = 6cm, ∠ DEB = 60 ° and find the chord CD length

Through o as of ⊥ CD in F, connect OC
∵AE=2CM,EB=6CM∴AB=8CM∴OA=OB=OC=4CM,OE=BE-OB=2CM
∵ DEB = 60 °, ofe = 90 °, EOF = 30 °, EF = 1 / 2oe = 1cm, Pythagorean theorem, of 2 = 3
In ⊿ OFC, Pythagorean theorem, CF = √ (4 ^ 2-3) = √ 13cm
According to the vertical diameter theorem, CD = 2cf = 2 √ 13cm

It is known that, as shown in the figure, the diameter ab of ⊙ o intersects the chord CD at the point E, AE = 1, be = 5, ∠ AEC = 45 ° and find the length of CD

Make oh ⊥ CD in H and connect OD, as shown in Fig,
∵AE=1，BE=5，
∴AB=AE+BE=6，
∴OA=OD=3，
∴OE=OA-AE=2，
∵∠AEC=45°，
∴OH=
Two
2OE=
2，
In RT △ ODH, DH=
OD2−OH2=
7，
∵OH⊥CD，
∴CH=DH=
7，
∴CD=2DH=2
7．

As shown in the figure, the diameter of circle O AB and chord CD intersect at point E, AE = 1cm, EB = 5cm, ∠ DEB = 30 ° and find the length of chord CD

Because AB is the diameter of a circle, so AB = AE + EB = 1 + 5 = 6 (CM), so Ao = 1 / 2Ab = 3, so OE = ao-ae = 3-1 = 2. In △ OEF, ∠ OEF = DEB = 30 ° so of = sin30 ° * OE = 1 / 2 * 2 = 1, so CD = 2cf = 2 * √ (3 ^ 2-1 ^

The diameter of circle O is ab = 12, chord CD is perpendicular to point E, and AE: EB = 1:3, CD =?

AE: EB = 1:3, indicating that AE is equal to 1 / 4 of diameter AB, AE = 3, OE = 3
In the right triangle OCE, OC = 6, according to Pythagorean theorem, CE = 3, radical 3 = 5.2
CD=2CE=10.4

As shown in the figure, the diameter AB and chord CD intersect at the point E, where AE = 1cm, EB = 5cm, ∠ DEB = 60 °, then the length of CD is___ .

Through point o as of ⊥ CD, connect OD,
∵AE=1cm，EB=5cm，
∴AB=AE+EB=1+5=6cm，
∴OA=OD=3cm，
∴OE=OA-AE=3-1=2cm，
In RT △ OEF, Deb = 60 ° and OE = 2cm,
∴OF=OE•sin∠DEB=2×
Three
2=
3cm，
In RT △ ODF,
DF=
OD2-OF2=
32-(
3)2=
6cm，
∵OF⊥CD，
∴CD=2DF=2×
6=2
6cm．
So the answer is: 2
6cm．

As shown in the figure, it is known that in ⊙ o, the chord ab ⊥ CD is e, AE = 2, EB = 8, and the degree of ⊙ CAD is 120 °, then the radius of ⊙ o is______ .

If OC, od are connected, O is used as of ⊥ AB in F, og ⊥ CD in G. then the quadrilateral ogef is a rectangle, og = EF
∵AE=2，EB=8，∴AB=10．
∵ of ⊥ AB in F, ᙽ AF = 1
2AB=5，∴EF=AF-AE=3=OG．
∵∠CAD=120°，∴∠COD=360°-2×120°=120°，
And ∵ OC = OD,  OCD = ∠ ODG = 30 °
∵ og ⊥ CD in G, ᙽ OC = 2og = 6
That is, the radius of ⊙ o is 6
So the answer is 6