It is known that, as shown in the figure, the diameter ab of the circle O intersects with the chord CD at the point E, AE = 1, be = 5, ∠ AEC = 30 ° and find the length of CD I can't draw a picture

Let CE = X
Make CF perpendicular to F
Then: Fe = CE / 2 = x / 2, CF ^ 2 = (3 / 4) x ^ 2
AF=1-(x/2)
FB=FE+EB=5+(x/2)
And: CF ^ 2 = af * EB
So: (3 / 4) x ^ 2 = [1 - (x / 2)] [5 + (x / 2)]
x^2+2x-5=0
X = (radical 6) - 1
However, CE * ed = AE * EB = 5
Ed = 5 / x = (radical 6) + 1
So: CD = CE + ed = 2 (radical 6)

Given that there are two strings AB = 8cm, CD = 6cm and ab ‖ CD in a circle of diameter 10, then the distance between strings AB and CD is______ .

It can be divided into two cases: when two chords are located on one side of the center O of the circle, as shown in Figure 1, through o as OE ⊥ CD, crossing CD at point E, crossing AB at point F, connecting OA, OC, ∵ ab ∥ CD, ᙽ OE ⊥ AB, ∵ e, f are the midpoint of CD and ab respectively,  CE = de = 12CD = 3cm, AF = BF = 12ab = 4cm, in RT △ AOF, OA = 5cm

Given that there are two strings AB = 8cm, CD = 6cm and ab ‖ CD in a circle of diameter 10, then the distance between strings AB and CD is______ .

The distance from the center of the circle to the center of the circle is known as CD = 12

Square of chord center distance = square of radius - square of half chord
The distance from O to AB is 10 square - (12 / 2) square = 8
The distance from O to CD is 10 square - (18 / 2) square = root 19
So the distance from the center of the circle to the intersection e of the two chords
=Under the root (8 square + 19 square)
=Radical 83

Find the distance between the two chords in the circle O with radius of 10cm AB / / CD and ab = 12, CD = 16

Since the longest chord of a point m passing through circle 0 is 10cm, it is known that the chord is diameter, so the radius r of ⊙ o is 5cm. Since the shortest chord passing through M should be vertically bisected by its diameter, OM ^ 2 = 5 ^ 2-4 ^ 2 = 9, so,

In the circle with radius 5, the chord AB is parallel to CD, AB is equal to 6cm, and CD is equal to 8cm. Find the distance between chord AB and CD? (drawing required)

The chord center distances of AB and CD are 4 and 3 respectively
When AB and CD are on the same side of the circle center, the distance between AB and CD is 4-3 = 1
When AB and CD are on the opposite side of the circle center, the distance between AB and CD is 4 + 3 = 7

If the radius of ⊙ o is 5cm, the chord ab ∥ CD, ab = 6cm, CD = 8cm, then the distance between AB and CD is () A. 1 cm B. 7 cm C. 1 cm or 7 cm D. It's impossible to judge

It can be divided into two cases: ① when AB and CD are on the same side of O, as shown in Fig. 1, OE ⊥ AB in E, Cd in F, OA, OC, ∵ ab ∥ CD, ∵ of ⊥ CD, ⊥ from the vertical diameter theorem, AE = 12ab = 3cm, CF = 12CD = 4cm, in RT △ OAE, from the Pythagorean theorem, OE = oa2 − AE2 = 52 − 32 = 4 (CM) is obtained

AB is the diameter of circle O, and the chord CD is perpendicular to E. if AB is equal to 10 cm and CD is equal to 8 cm, what is the length of OE

25-16 = 9 answer = 3

Given that AB and CD are two parallel chords of circle O, and ab = 48, CD = 40, and the distance between the two parallel chords is 22, what is the radius of circle O

AB to the center of the circle is X
x^2 + 24^2 = (x+22)^2 + 20^2 = R^2
576 = 44x + 884 x = -7
R=25
7,24,25 15,20,25

It is known that, as shown in the figure, the diameter ab of the circle O intersects with the chord CD at the point E, AE = 1, be = 5, ∠ AEC = 30 ° and find the length of CD I can't draw a picture