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

In circle O, the chords AB and CD intersect perpendicularly at point E, AE = 5, CE = 1, be = 3

In the theorem circle, if AB is perpendicular to CD, there is AE times be = CE times De = 15

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．

In circle O, if chord CD intersects diameter AB at point E, angle AEC = 30 degrees AE = 1 be = 5, then chord center distance of chord CD is of=__ String CD=_ In circle O, if chord CD intersects diameter AB at point E, angle AEC = 30 degrees AE = 1 be = 5, then chord center distance of chord CD is of=__ String CD=_ kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk The solution should be specific

Of = 1, CD = 4 times root 2
Because:
AB=AE+BE=6
OA=3
OE=OA-AE=2
Angle AEC = 30 degrees
So, of = 1 / 2 * OE = 1
CF^2=OC^2-OE^2=3^2-1^2=8
CF = 2 times root 2
CD = 2 * CF = 4 times root 2

AB is the diameter of circle O, the chord CD intersects AB at point E, AE = 3, be = 5, ∠ AEC = 30 ° to find the length of CD

Oh ⊥ CD
The radius of the circle O is 4
∴OE=1
In right triangle EOH
And ? CEA = ∵ OEH = 30 
∴2OH=OE
OH=0.5
In the right triangle OHC
Using Pythagorean theorem, ch = (3 √ 7) / 2 is calculated
According to the vertical diameter theorem, CD = 3 √ 7

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

The diameter of circle O is 10cm, the chord AB is parallel to CD and a B is equal to 6cm, and DC is equal to 8cm. Find the distance between AB and CD. Be careful

There are two distances, one on one side and one on both sides of the center
The distance from chord AB to the center of the circle is 4, and the distance from string CD to the center of the circle is 3,
So if it's on the side of the center of the circle, the distance is 1
If on both sides of the center of the circle, the distance is 7

Given the chord AB = 6cm, diameter CD = 10cm of circle O, and ab is perpendicular to CD, what is the distance between point C and ab?

AB vertical CD, vertical foot e, connected Ao
AO=CO=CD/2=10/2=5,
AE=AB/2=6/2=3,
EO²=AO²-AE²=5²-3²=4²
EO=3cm；
The distance from point C to AB is 3cm

It is known that the radius of circle O is 10cm, the chord AB is parallel to CD, ab = 6, CD = 8. Find the distance between AB and CD

It is known that the radius of circle O is r = 10cm, the chord AB is parallel to CD, ab = 6, CD = 8. Find the distance between AB and CD
Ab.. Distance between CdS = (R ^ 2 - (AB / 2) ^ 2) ^ 0.5 - (R ^ 2 - (CD / 2) ^ 0.5
=(10^2-(6/2)^2)^0.5-(10^2-(8/2)^2)^0.5
=0.374cm

The radius of circle O is 5cm, the chord AB is equal to the chord CD, AB is equal to 6cm, and CD is equal to 8cm. Find the distance between AB and CD

It's string ab ‖ string CD,
According to Pythagorean theorem,
The distance from point O to AB is √ (5 ^ 2-3 ^ 2) = 4 cm,
The distance from point O to CD is √ (5 ^ 2-4 ^ 2) = 3 cm,
When AB and CD are on the same side of the circle center, the distance between AB and CD is 4-3 = 1 cm,
When AB and CD are on both sides of the circle center, the distance between AB and CD is 4 + 3 = 7 cm

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______ .