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

If the distance from O to AB is x, then the distance from O to CD is 22-x
that
24²+x²=20²+(22-x)²
The solution
X=7
R²=7²+24²=625
R=25
The radius of circle O is 25

The diameter of circle O is 50 cm, the chord AB is parallel to CD, and ab is equal to 40 cm, and CD is equal to 48 cm I want the right answer. I need it very urgently. Please reply me as soon as possible,

There are two answers: 22cm or 8cm
If CD and ab are on the same side of the graph, then CD and ab are on the same side
According to the definition of circle, EO is perpendicular to the diameter, fo is perpendicular to the diameter, and EOF is on the same line
Because AO and Co are the radius of a circle, so Ao = co = 25cm, and because EO bisects AB and fo equally, so AE = 20cm, CF = 24cm
From the theorem, we can get: EO = 25 square minus 20 square, then open square = 15 cm, the same reason fo = 7 cm
EF = 15 plus or minus 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

In a circle with a radius of 5cm, chord AB = 6, chord CD = 8, and ab ‖ CD, find the distance between AB and CD

Make a straight line perpendicular to the chord, where EF intersects AB, e intersects CD, and f connects AE CF
In the 3-intersection a0e, the Pythagorean theorem gives E0 = 4
In the 3-angle c0f, the Pythagorean theorem leads to F0 = 3
So the distance of AB CD should be 4-3 = 1 and 4 + 3 = 7 (in the case of a semicircle and not in a semicircle)
Test site: two situations should be considered

As shown in the figure, the diameter ab of ⊙ o is 10cm, the chord AC is 6cm, the bisector of ⊙ ACB intersects ⊙ o in D, and find the length of BC, ad, BD

∵ AB is the diameter
∴∠ACB=∠ADB=90°
In RT △ ABC, AB2 = ac2 + BC2, ab = 10cm, AC = 6cm
∴BC2=AB2-AC2=102-62=64
∴BC=
64=8（cm）
And CD bisection ∠ ACB,
∴∠ACD=∠BCD，
Qi
AD＝
DB
∴AD=BD
In RT △ abd, ad2 + BD2 = AB2
∴AD2+BD2=102
∴AD=BD=
One hundred
2=5
2（cm）．

As shown in the figure, the diameter ab of ⊙ o is 10cm, the chord AC is 6cm, the bisector of ⊙ ACB intersects ⊙ o in D, and find the length of BC, ad, BD

 AB is the diameter  ACB = ∠ ADB = 90 ° in RT △ ABC, AB2 = ac2 + BC2, ab = 10cm, AC = 6cm ᙽ BC2 = ab2-ac2 = 102-62 = 64 ᙽ BC = 64 = 8 (CM) and CD bisection

In this paper, we find the length of the graph of ⊙ a, a, B, a, B, a, B, a, B, a, B, a, B, a, B, a, B, a, B, a, B, a, B, a, B, a, B, a, B, a, B, a, B, a, B, a, B, a, B, B, a, B, B, a

1. (Fig. 1)
(1) AB is the diameter
﹤ ACB is a right angle (the circular angle on a semicircle is a right angle)
By using Pythagorean theorem, we can obtain the following results:
BC=8
(2) ∵ CD bisection ∵ ACB
∴∠ACD=∠BCD=90°÷2=45°
In the same circle, the angle of the circle opposite to the arc is equal
Similarly, ∠ abd = ∠ ACD
∴AD=BD
And ∠ ACB is a right angle
It can be obtained by Pythagorean theorem
Ad = 5 √ 2 (5 pieces, 2 pieces)
2. (Fig. 2)
∵OC⊥AP OD⊥BP
/ / AC = CP, PD = DP (the diameter perpendicular to the chord bisects the chord)
The CD is the median line of △ ABP
ν CD = 1 / 2Ab = 1 / 2 × 8 = 4f (the median line of a triangle is equal to half of the length of the bottom edge)

The diameter EF is 10cm, chord AB, CD are 6cm8cm, AB / / EF / / CD Connect AE, be, FD, FC to find the shadow enclosed by AE, AB and 0 arc AB and the sum of shadow area enclosed by FC, FD and arc CD The process should be detailed

The distance from AB to EF is 4 (Pythagorean theorem, 3, 4, 5)
The distance from CD to EF is 3 (same as above)
The area obtained is the sum of the areas of sector OAB and sector OCD (o is the center of the circle) -
A small arc and a triangle form a sector. The area of triangle OAB and triangle EAB are the same because of the same bottom and same height
The angle of the two sectors is 2x (arcsin (3 / 5) + arcsin (4 / 5)) = 180 degrees
So the area is the area of the half circle, 3.14x5x5 / 2 = 39.25cm square

As shown in the figure: the diameter of ⊙ o is 10cm, the chord AB is 8cm, and P is the point on chord ab. if the length of OP is an integer, then the point P satisfying the condition is______ One

OC ⊥ AB is used as OC ⊥ AB is connected to OA through o;
In RT △ OAC, OA = 5cm, AC = 4cm;
∴OC=
OA2−AC2=3cm；
∴3≤OP≤5；
Therefore, Op = 3cm, 4cm, or 5cm;
When OP = 3cm, point P coincides with point C, and there is a qualified point P;
When OP = 4cm, P is between AC and BC, and there are two p points that meet the conditions;
When OP = 5cm, P coincides with a or B, and there are two qualified P points;
Therefore, there are five p points satisfying the condition

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